Integrate the following function: (1)/(cos² x (1 - – Class 12 Mathematics | ConceptFirst

Exercise 7.2 · Q25

\frac{1}{\cos^2 x (1 - \tan x)^2}

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Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The chain rule differentiates composite functions: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given $\cos(x^2) \cdot 2x$ , find the original function. That's what u substitution does — it reverses the chain rule.

The Core Intuition

When an integral looks like "a function times the derivative of its inside," substitute the inside with $u$ and the derivative of the inside with $du$ . Consider:

\int 2x \cos(x^2) \, dx

Here $2x$ is the derivative of $x^2$ , and $x^2$ is the inside of $\cos(x^2)$ . Let $u = x^2$ , so $du = 2x \, dx$ :

\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C

Check: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ .

The Precise Statement

\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x), \, du = g'(x) \, dx

Valid provided $g$ is differentiable and the resulting integral in $u$ is simpler.

The Step-by-Step Method

Identify a function $g(x)$ whose derivative $g'(x)$ also appears (possibly up to a constant factor).
Set $u = g(x)$ , compute $du = g'(x) \, dx$ .
Rewrite the entire integral in $u$ and $du$ — every $x$ and $dx$ must be replaced.
Integrate with respect to $u$ .
Substitute back $u = g(x)$ .

Watch out

You cannot mix variables. If any $x$ remains after substitution, you chose the wrong $u$ (or must solve for $x$ in terms of $u$ — rare).

A Second Example (with a constant factor)

Evaluate $\int x \sqrt{x^2 + 1} \, dx$ . Let $u = x^2 + 1$ , so $x \, dx = \frac{1}{2} du$ :

\int \sqrt{u} \cdot \tfrac{1}{2} du = \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} + C = \tfrac{1}{3} (x^2 + 1)^{3/2} + C

When Does It Work?

When the integrand is something times the derivative of something inside. Common patterns:

$x \cdot f(x^2)$ — derivative of $x^2$ is $2x$ , so $u = x^2$
$e^{g(x)} \cdot g'(x)$ — derivative of $g(x)$ appears
$\frac{g'(x)}{g(x)}$ — leads to $\log|g(x)|$ …

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With $u = \tan x$ , $\displaystyle\int \frac{dx}{\cos^2 x\,(1-\tan x)^2} = \frac{1}{1-\tan x} + C.$

1. Use $\dfrac{1}{\cos^2 x} = \sec^2 x$ .

\int \frac{\sec^2 x}{(1-\tan x)^2}\,dx.

2. Substitute. Let $u = \tan x$ , so $du = \sec^2 x\,dx$ :

\int \frac{du}{(1-u)^2}.

3. Integrate.

$\int (1-u)^{-2}\,du = \frac{1}{1-u} + C,$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 51 on this concept.

AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$\displaystyle\int \dfrac{1}{\left(\sin x + \cos x + \sqrt{2}\sqrt{\sin 2x}\right)^2}\,dx =$
›Reveal solutionSolution
Recognising the denominator as $(\sqrt{\sin x}+\sqrt{\cos x})^4$ and substituting $u=\sqrt{\tan x}$ reduces this to a rational integral, giving $-\dfrac{1+3\sqrt{\tan x}}{3(1+\sqrt{\tan x})^3}+C$ .

Concept and Intuition

The key algebraic identity here is $(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x + \cos x + 2\sqrt{\sin x\cos x} = \sin x+\cos x+\sqrt{2\sin 2x}$ (since $2\sqrt{\sin x\cos x}=\sqrt{4\sin x\cos x}=\sqrt{2\sin2x}$ ). That matches the given denominator's base exactly, turning a scary-looking radical expression into a clean fourth power.

Step-by-Step Solution

Verify $(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x+\cos x+2\sqrt{\sin x\cos x} = \sin x+\cos x+\sqrt{2\sin2x}$ , matching $\sin x+\cos x+\sqrt2\sqrt{\sin2x}$ .

So the denominator is $(\sqrt{\sin x}+\sqrt{\cos x})^4$ .

Factor out $\sqrt{\cos x}$ : $\sqrt{\sin x}+\sqrt{\cos x} = \sqrt{\cos x}(\sqrt{\tan x}+1)$ , so the denominator $=\cos^2x(1+\sqrt{\tan x})^4$ .

Integral becomes $\displaystyle\int \dfrac{\sec^2x\,dx}{(1+\sqrt{\tan x})^4}$ . Let $t=\tan x$ , $dt=\sec^2x\,dx$ : $\displaystyle\int \dfrac{dt}{(1+\sqrt t)^4}$ .

Let $u=\sqrt t$ , $t=u^2$ , $dt=2u\,du$ : $\displaystyle\int \dfrac{2u\,du}{(1+u)^4}$ .

Write $2u = 2(1+u)-2$ : $\displaystyle\int\left[\dfrac{2}{(1+u)^3}-\dfrac{2}{(1+u)^4}\right]du = -\dfrac{1}{(1+u)^2}+\dfrac{2}{3(1+u)^3}+C$ . …
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AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$\displaystyle\int \dfrac{\sec^2 x}{(\sec x+\tan x)^{5/2}}\,dx=$
›Reveal solutionSolution
A substitution $t=\sec x+\tan x$ (which pairs neatly with $\sec x-\tan x=1/t$ ) reduces this odd-looking integral to a simple power-rule integral, whose answer re-expresses in terms of $\sec x-\tan x$ .

Concept and Intuition

Whenever $\sec x+\tan x$ appears, remember its reciprocal identity $(\sec x+\tan x)(\sec x-\tan x)=1$ , and that $\dfrac{d}{dx}(\sec x+\tan x)=\sec x(\sec x+\tan x)$ — this makes $t=\sec x+\tan x$ a natural substitution whenever $\sec x\,dx$ multiplies a function of $t$ .

Step-by-Step Solution

Let $t=\sec x+\tan x$ . Then $dt=\sec x(\sec x+\tan x)\,dx=\sec x\cdot t\,dx$ , so $\sec x\,dx=\dfrac{dt}{t}$ .

Also $\sec x-\tan x=\dfrac1t$ , so $\sec x=\dfrac{t+1/t}{2}=\dfrac{t^2+1}{2t}$ .

The integral $\displaystyle\int\frac{\sec^2x}{t^{5/2}}dx=\int\frac{\sec x}{t^{5/2}}\cdot(\sec x\,dx)=\int\frac{\sec x}{t^{5/2}}\cdot\frac{dt}{t}=\int\frac{\sec x}{t^{7/2}}dt$ .

Substitute $\sec x=\dfrac{t^2+1}{2t}$ : integral $=\displaystyle\int\frac{t^2+1}{2t^{9/2}}dt=\frac12\int\left(t^{-5/2}+t^{-9/2}\right)dt$ .

$=\dfrac12\left[-\dfrac{2}{3}t^{-3/2}-\dfrac27 t^{-7/2}\right]+c=-\dfrac{t^{-3/2}}{3}-\dfrac{t^{-7/2}}{7}+c$ . …
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AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$\displaystyle\int \dfrac{\sin^2 x \cos^2 x}{\cos^6 x + \sin^6 x}\,dx =$
›Reveal solutionSolution
Dividing through by $\cos^6x$ to introduce $\tan x$ , then a double substitution ( $t=\tan x$ , then $u=t^3$ ), reduces the integral to a standard arctangent form, giving $\dfrac13\mathrm{Tan}^{-1}(\tan^3x)+c$ .

Concept and Intuition

When an integrand has $\sin$ and $\cos$ appearing only in even powers that can be grouped, dividing everything by the highest power of $\cos x$ converts the whole expression into a rational function purely of $\tan x$ — a very common trick that opens the door to the substitution $t=\tan x$ . Here, since the denominator naturally forms $1+\tan^6x$ , a further substitution $u=t^3$ turns it into the standard $\int\frac{du}{1+u^2}$ arctangent integral.

Step-by-Step Solution

Divide numerator and denominator of $\dfrac{\sin^2x\cos^2x}{\cos^6x+\sin^6x}$ by $\cos^6x$ : numerator becomes $\dfrac{\sin^2x\cos^2x}{\cos^6x} = \dfrac{\sin^2x}{\cos^4x} = \tan^2x\sec^2x$ ; denominator becomes $1 + \tan^6x$ .

The integral is now $\displaystyle\int \dfrac{\tan^2x\sec^2x}{1+\tan^6x}\,dx$ .

Substitute $t=\tan x$ , so $dt = \sec^2x\,dx$ . The integral becomes $\displaystyle\int\dfrac{t^2}{1+t^6}\,dt$ .

Substitute $u = t^3$ , so $du = 3t^2\,dt$ , i.e. $t^2\,dt = \dfrac{du}{3}$ . The integral becomes $\dfrac13\displaystyle\int\dfrac{du}{1+u^2}$ . …
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AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$\displaystyle\int(\tan^7x+\tan x)\,dx=$
›Reveal solutionSolution
Factoring $\tan^7x+\tan x$ using the sum-of-like-terms identity $t^7+t=t(t^6+1)=t(t^2+1)(t^4-t^2+1)$ exposes the $\sec^2x$ needed for a clean $t=\tan x$ substitution. Answer: $\dfrac{\tan^2x}{12}(2\tan^4x-3\tan^2x+6)+c$ .

Concept and Intuition

Whenever an integrand is built purely from powers of $\tan x$ together with an explicit or hidden $\sec^2x$ , substituting $t=\tan x$ turns it into a polynomial integral — the key is recognizing that $t^6+1=(t^2+1)(t^4-t^2+1)$ supplies exactly the $\sec^2x=1+\tan^2x$ factor needed.

Step-by-Step Solution

$\tan^7x+\tan x=\tan x(\tan^6x+1)=\tan x(\tan^2x+1)(\tan^4x-\tan^2x+1)=\tan x\sec^2x(\tan^4x-\tan^2x+1)$ .

Let $t=\tan x$ , $dt=\sec^2x\,dx$ . Integral becomes $\displaystyle\int t(t^4-t^2+1)dt=\int(t^5-t^3+t)dt$ .

$=\dfrac{t^6}{6}-\dfrac{t^4}{4}+\dfrac{t^2}{2}+c=\dfrac{\tan^6x}{6}-\dfrac{\tan^4x}{4}+\dfrac{\tan^2x}{2}+c$ . …
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AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
$\displaystyle\int \dfrac{dx}{\left(1+\sqrt{x}\right)^{2022}} =$
›Reveal solutionSolution
Substituting $t=1+\sqrt x$ turns the integral into a simple power-rule integral in $t$ ; back-substituting and factoring reproduces option (A)'s bracketed form.

Concept and Intuition

Whenever an integrand is a function purely of $1+\sqrt x$ , the substitution $t=1+\sqrt x$ (so $\sqrt x = t-1$ , $x=(t-1)^2$ ) turns the messy radical expression into a clean power of $t$ , and the pieces of $dx$ that are left over ( $2(t-1)\,dt$ ) combine with the $t^{-2022}$ factor to give a difference of two pure power terms — each integrable by the ordinary power rule.

Step-by-Step Solution

Let $t=1+\sqrt x$ . Then $\sqrt x=t-1$ , $x=(t-1)^2$ , and $dx=2(t-1)\,dt$ .

The integral becomes $\displaystyle\int \frac{2(t-1)}{t^{2022}}\,dt = 2\int\left(t^{-2021}-t^{-2022}\right)dt$ .

Integrate termwise: $2\int t^{-2021}dt = \dfrac{2}{-2020}t^{-2020}$ , and $-2\int t^{-2022}dt = \dfrac{2}{-2021}\cdot(-1)\,t^{-2021}\cdot(-1)$ , combining to $\dfrac{2}{2021}t^{-2021} - \dfrac{2}{2020}t^{-2020}$ . …
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AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
$\displaystyle\int \dfrac{dx}{4+5\cos x} =$
›Reveal solutionSolution
The Weierstrass substitution $t=\tan(x/2)$ turns this into a standard $\int dt/(a^2-t^2)$ integral, giving option (B).

Concept and Intuition

For $\int\dfrac{dx}{a+b\cos x}$ the substitution $t=\tan(x/2)$ (so $\cos x=\tfrac{1-t^2}{1+t^2}$ , $dx=\tfrac{2\,dt}{1+t^2}$ ) always converts the integral into a rational function of $t$ alone.

Step-by-Step Solution

$\cos x=\dfrac{1-t^2}{1+t^2}$ , $dx=\dfrac{2\,dt}{1+t^2}$ , with $t=\tan(x/2)$ .

$4+5\cos x=4+5\cdot\dfrac{1-t^2}{1+t^2}=\dfrac{4(1+t^2)+5(1-t^2)}{1+t^2}=\dfrac{9-t^2}{1+t^2}$ .

Integral becomes $\int\dfrac{2\,dt/(1+t^2)}{(9-t^2)/(1+t^2)}=\int\dfrac{2\,dt}{9-t^2}$ .

Using $\int\dfrac{dt}{a^2-t^2}=\dfrac{1}{2a}\log\left|\dfrac{a+t}{a-t}\right|+C$ with $a=3$ : $\int\dfrac{dt}{9-t^2}=\dfrac16\log\left|\dfrac{3+t}{3-t}\right|+C$ . …
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AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$f(x) = \dfrac{x}{(1 + nx^n)^{1/n}}$
›Reveal solutionSolution
A direct substitution $u=1+nx^n$ turns the integral into a simple power rule.

Concept and Intuition

The integrand's power of $x$ (namely $x^{n-1}$ ) is exactly proportional to the derivative of $u=1+nx^n$ , which is the classic signal to substitute.

Step-by-Step Solution

$x^{n-2}f(x)=x^{n-2}\cdot\dfrac{x}{(1+nx^n)^{1/n}}=\dfrac{x^{n-1}}{(1+nx^n)^{1/n}}$ .

Let $u=1+nx^n$ . Then $du=n\cdot n x^{n-1}dx=n^2x^{n-1}dx$ , so $x^{n-1}dx=\dfrac{du}{n^2}$ .

Integral becomes $\displaystyle\int u^{-1/n}\cdot\dfrac{du}{n^2}=\dfrac{1}{n^2}\cdot\dfrac{u^{1-1/n}}{1-1/n}+C$ .

Simplify: $\dfrac{1}{n^2}\cdot\dfrac{1}{\frac{n-1}{n}}=\dfrac{1}{n^2}\cdot\dfrac{n}{n-1}=\dfrac{1}{n(n-1)}$ . …
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AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$\displaystyle\int \dfrac{dx}{\sqrt{\sin^3 x \cos(x-\alpha)}} =$
›Reveal solutionSolution
Expanding $\cos(x-\alpha)$ and substituting $u=\cot x$ reduces this to a simple square-root integral; final answer is (D).

Concept and Intuition

The key move is expanding $\cos(x-\alpha)$ and factoring out $\sin^4x$ from inside the square root so that a substitution $u=\cot x$ (whose differential is $-\csc^2x\,dx$ , conveniently matching what's left outside) linearizes the whole thing.

Step-by-Step Solution

$\cos(x-\alpha)=\cos x\cos\alpha+\sin x\sin\alpha$ .

$\sin^3x\cos(x-\alpha)=\sin^3x\cos x\cos\alpha+\sin^4x\sin\alpha = \sin^4x\left(\dfrac{\cos x}{\sin x}\cos\alpha+\sin\alpha\right)=\sin^4x(\cot x\cos\alpha+\sin\alpha)$ .

So $\sqrt{\sin^3x\cos(x-\alpha)}=\sin^2x\sqrt{\cot x\cos\alpha+\sin\alpha}$ (taking $\sin^2x>0$ outside the root).

The integral is $\displaystyle\int\dfrac{dx}{\sin^2x\sqrt{\cos\alpha\cot x+\sin\alpha}}=\int\dfrac{\csc^2x\,dx}{\sqrt{\cos\alpha\cot x+\sin\alpha}}$ .

Let $u=\cot x$ , $du=-\csc^2x\,dx$ . Integral becomes $-\displaystyle\int\dfrac{du}{\sqrt{\cos\alpha\,u+\sin\alpha}}=-\dfrac{2\sqrt{\cos\alpha\,u+\sin\alpha}}{\cos\alpha}+c$ . …
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AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$\displaystyle\int \dfrac{1}{\sin^3 x+\cos^3 x}\,dx = A\log\left|\dfrac{\sqrt{2}+t}{\sqrt{2}-t}\right|+B\,\mathrm{Tan}^{-1}(t)+c$
›Reveal solutionSolution
Factoring the sum of cubes and substituting $t=\sin x-\cos x$ turns the trigonometric integral into a clean rational-function integral in $t$ , giving $B/A=2\sqrt2$ with $t=\sin x-\cos x$ .

Concept and Intuition

$\sin^3x+\cos^3x$ factors as a sum of cubes: $(\sin x+\cos x)(1-\sin x\cos x)$ . Both $\sin x+\cos x$ and $1-\sin x\cos x$ can be written purely in terms of $u=\sin x-\cos x$ , because $(\sin x+\cos x)^2+(\sin x-\cos x)^2=2$ and $1-\sin x\cos x=\tfrac{1+(\sin x-\cos x)^2}{2}$ . Crucially, $\dfrac{du}{dx}=\cos x+\sin x$ , which is exactly the factor left over after using the second identity — so the whole integral collapses into a rational function of $u$ alone.

Step-by-Step Solution

Factor: $\sin^3x+\cos^3x=(\sin x+\cos x)(1-\sin x\cos x)$ .

Let $u=\sin x-\cos x$ . Then $u^2=1-2\sin x\cos x$ , so $1-\sin x\cos x=\dfrac{1+u^2}{2}$ .

Also $(\sin x+\cos x)^2=1+2\sin x\cos x=2-u^2$ , so $\sin x+\cos x=\sqrt{2-u^2}$ (taking the appropriate branch), and $\dfrac{du}{dx}=\cos x+\sin x=\sqrt{2-u^2}$ .

So $\sin^3x+\cos^3x=\sqrt{2-u^2}\cdot\dfrac{1+u^2}{2}$ , and

$I=\int\frac{dx}{\sin^3x+\cos^3x}=\int\frac{2\,dx}{\sqrt{2-u^2}(1+u^2)}=\int\frac{2}{\sqrt{2-u^2}(1+u^2)}\cdot\frac{du}{\sqrt{2-u^2}}=\int\frac{2\,du}{(2-u^2)(1+u^2)}.$

Partial fractions (by symmetry, only even terms survive): $\dfrac{2}{(2-u^2)(1+u^2)}=\dfrac{2/3}{2-u^2}+\dfrac{2/3}{1+u^2}$ . …
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AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
$\displaystyle\int \sqrt{x+\sqrt{x^2+2}}\,dx =$
›Reveal solutionSolution
The substitution $t=x+\sqrt{x^2+2}$ rationalises the nested radical; the resulting antiderivative, written as a single fraction, matches option (D). Answer: option (D).

Concept and Intuition

Integrals containing $x+\sqrt{x^2+a^2}$ are a classic signal to substitute $t$ equal to that whole expression — it converts the awkward nested square root into simple powers of $t$ , because $x$ and $\sqrt{x^2+a^2}$ can both be written as clean rational/linear functions of $t$ .

Step-by-Step Solution

Let $t=x+\sqrt{x^2+2}$ . Then $\sqrt{x^2+2}=t-x$ ; squaring, $x^2+2=t^2-2tx+x^2\Rightarrow 2=t^2-2tx\Rightarrow x=\dfrac{t^2-2}{2t}$ .

Then $\sqrt{x^2+2}=t-x=t-\dfrac{t^2-2}{2t}=\dfrac{2t^2-t^2+2}{2t}=\dfrac{t^2+2}{2t}$ .

Differentiate $t$ w.r.t. $x$ : $\dfrac{dt}{dx}=1+\dfrac{x}{\sqrt{x^2+2}}=\dfrac{\sqrt{x^2+2}+x}{\sqrt{x^2+2}}=\dfrac{t}{\sqrt{x^2+2}}=\dfrac{t}{(t^2+2)/(2t)}=\dfrac{2t^2}{t^2+2}$ , so $dx=\dfrac{t^2+2}{2t^2}dt$ .

Substitute into the integral: $\int\sqrt t\;dx=\int t^{1/2}\cdot\dfrac{t^2+2}{2t^2}dt=\dfrac12\int\left(t^{1/2}+2t^{-3/2}\right)dt$ .

Integrate: $\dfrac12\left[\dfrac23t^{3/2}-4t^{-1/2}\right]+C=\dfrac13t^{3/2}-2t^{-1/2}+C$ . …
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AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$\displaystyle\int \dfrac{\cos^3 x}{(1 + \sin x)^4}\, dx =$
›Reveal solutionSolution
Factor $\cos^3x$ using $\cos^2x=(1-\sin x)(1+\sin x)$ and substitute $t=\sin x$ ; the resulting antiderivative can equivalently be written in the $\cos^4x/(1+\sin x)^4$ form given in the options (they differ only by an added constant). Answer: $-\dfrac{\cos^4x}{4(1+\sin x)^4}+c$ .

Concept and Intuition

Integrals of $\cos^{\text{odd}}x$ over powers of $(1+\sin x)$ are handled by peeling off one factor of $\cos x$ to pair with $dx$ (making $d(\sin x)$ ) and expressing the remaining even power of $\cos x$ in terms of $\sin x$ . Since the MCQ options are phrased in terms of $\cos^4x$ rather than $\sin x$ directly, it is often faster (and safer against sign traps) to guess-and-check an antiderivative of that shape by differentiating a general form $A\cos^4x(1+\sin x)^{-n}$ and matching powers/coefficients — this is exactly how the printed option is confirmed.

Step-by-Step Solution

$\cos^3x=\cos x\cdot\cos^2x=\cos x(1-\sin^2x)=\cos x(1-\sin x)(1+\sin x)$ .

Integrand $=\dfrac{\cos x(1-\sin x)(1+\sin x)}{(1+\sin x)^4}=\dfrac{\cos x(1-\sin x)}{(1+\sin x)^3}$ .

Let $t=\sin x$ , $dt=\cos x\,dx$ : $I=\displaystyle\int\frac{1-t}{(1+t)^3}dt$ . Writing $1-t=2-(1+t)$ : $I=\displaystyle\int\left(\frac{2}{(1+t)^3}-\frac{1}{(1+t)^2}\right)dt=-\frac{1}{(1+t)^2}+\frac{1}{1+t}+c=\frac{t}{(1+t)^2}+c$ .

So $I=\dfrac{\sin x}{(1+\sin x)^2}+c$ is one valid closed form. …
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AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
$\displaystyle\int \dfrac{\sin^6 x}{\cos^8 x}\,dx =$
›Reveal solutionSolution
$\int \tan^6x\sec^2x\,dx = \dfrac{\tan^7x}{7}+c$ via $u=\tan x$ .

Concept and Intuition

$\sin^6x/\cos^8x = \tan^6x\cdot\sec^2x$ , and $\sec^2x\,dx$ is exactly $d(\tan x)$ — a direct power-rule substitution.

Step-by-Step Solution

$\dfrac{\sin^6x}{\cos^8x} = \tan^6x\cdot\dfrac{1}{\cos^2x} = \tan^6x\sec^2x$ .

Let $u=\tan x \Rightarrow du=\sec^2x\,dx$ .

$\int u^6\,du = \dfrac{u^7}{7}+c = \dfrac{\tan^7x}{7}+c$ .

Common Mistakes …
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