(a) Find : ∫ (cos x)/((4+sin² x)(5-4cos² x)) dx OR (b) Evaluate : ∫₀^π (dx)/(a² cos² x + b² sin² x)

Question

\int \frac{\cos x}{(4+\sin^2 x)(5-4\cos^2 x)}\, dx

(OR)

(b) Evaluate :

\int_0^\pi \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}

CBSECBSE Class XII Board 2025Subjective· 5mImportance★★★★★

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Part (a)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 $\log|g(x)|$ …

Part (b)Concept understanding — Definite Substitution Method

Substitution in Definite Integrals

You already know substitution for indefinite integrals: set $u = g(x)$ , rewrite in terms of $u$ , integrate, then substitute back. For a definite integral there is a cleaner twist — instead of substituting back, you convert the limits of integration to the new variable and finish entirely in $u$ .

Why the limits must change

The limits $a$ and $b$ are $x$ -values. Once you switch to $u = g(x)$ , those numbers no longer describe the start and end of the integration — the corresponding $u$ -values do. Keeping the old numbers would integrate over the wrong interval, like reading a distance in kilometres off a scale marked in miles.

\int_{a}^{b} f\big(g(x)\big)\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du

The steps

Choose $u = g(x)$ , picking something whose derivative already appears in the integrand.
Differentiate: $du = g'(x)\,dx$ .
Convert the limits: the lower limit becomes $u = g(a)$ , the upper becomes $u = g(b)$ .
Integrate in $u$ — no substituting back needed.

Example. Evaluate $\int_0^2 2x\,(x^2+1)^3\,dx$ .

Let $u = x^2 + 1$ , so $du = 2x\,dx$ . When $x=0$ , $u=1$ ; when $x=2$ , $u=5$ . Then

\int_0^2 (x^2+1)^3\,(2x\,dx) = \int_1^5 u^3\,du = \left[\tfrac{u^4}{4}\right]_1^5 = \tfrac{625-1}{4} = 156.

We never returned to $x$ — the converted limits carried the work. …

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With $t=\sin x$ the integral becomes $\int\frac{dt}{(4+t^2)(1+4t^2)}$ ; partial fractions give $-\tfrac1{30}\tan^{-1}\frac{\sin x}{2}+\tfrac{2}{15}\tan^{-1}(2\sin x)+C$ .
Using symmetry and $t=\tan x$ , the value is $\dfrac{\pi}{ab}$ .

Part (a)

Reduce to one variable. Since $\cos^2x=1-\sin^2x$ ,

5-4\cos^2x=5-4(1-\sin^2x)=1+4\sin^2x.

With the $\cos x\,dx$ in the numerator, substitute $t=\sin x$ , $dt=\cos x\,dx$ :

I=\int\frac{dt}{(4+t^2)(1+4t^2)}.

Partial fractions. Both factors are even in $t$ , so try

\frac{1}{(4+t^2)(1+4t^2)}=\frac{A}{4+t^2}+\frac{B}{1+4t^2}\ \Rightarrow\ 1=A(1+4t^2)+B(4+t^2).

Comparing coefficients: $4A+B=0$ and $A+4B=1$ . Solving, $A=-\tfrac1{15},\ B=\tfrac4{15}$ , so

\frac{1}{(4+t^2)(1+4t^2)}=\frac{-1/15}{4+t^2}+\frac{4/15}{1+4t^2}.

Integrate. Using $\int\frac{dt}{a^2+t^2}=\frac1a\tan^{-1}\frac ta$ and $\int\frac{dt}{1+4t^2}=\tfrac12\tan^{-1}(2t)$ :

I=-\frac{1}{15}\cdot\frac12\tan^{-1}\frac t2+\frac{4}{15}\cdot\frac12\tan^{-1}(2t)+C=-\frac{1}{30}\tan^{-1}\frac t2+\frac{2}{15}\tan^{-1}(2t)+C.

Back-substitute $t=\sin x$ . …

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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.

CBSE 2026Set 65/1/11 markMCQ
$\int \frac{3ax}{b^2 + c^2x^2} dx = A \log |b^2 + c^2x^2| + K$
›Reveal solutionSolution
The integral fits the pattern $\int \frac{du}{u} = \log|u| + C$ after a substitution. The constant $A$ turns out to be $\frac{3a}{2c^2}$ , which corresponds to option (D).

The problem gives you the result of an integral and asks you to identify the constant $A$ that makes the equation true. This is a classic "match the form" question — you don't need to guess; you just need to perform the integration carefully and compare.

The key insight is that the integrand $\frac{3ax}{b^2 + c^2x^2}$ is a rational function where the numerator is almost the derivative of the denominator. The derivative of $b^2 + c^2x^2$ is $2c^2x$ . Our numerator is $3ax$ , which is a constant multiple of $x$ . So a simple substitution $u = b^2 + c^2x^2$ will turn the integral into $\int \frac{du}{u}$ .

Let's work through it step by step.

Set up the substitution.

Let $u = b^2 + c^2x^2$ . Then $du = 2c^2x \, dx$ , so $x \, dx = \frac{du}{2c^2}$ .

Rewrite the integral in terms of $u$ .

The integral is $\int \frac{3ax}{b^2 + c^2x^2} \, dx = \int \frac{3a}{u} \cdot (x \, dx)$ .

Substitute $x \, dx = \frac{du}{2c^2}$ :

$\int \frac{3a}{u} \cdot \frac{du}{2c^2} = \frac{3a}{2c^2} \int \frac{du}{u}.$

Integrate. $\int \frac{du}{u} = \log|u| + C$ , so

$\frac{3a}{2c^2} \log|u| + C = \frac{3a}{2c^2} \log|b^2 + c^2x^2| + K,$
where $K$ is the constant of integration (we renamed $C$ to $K$ to match the problem).

Compare with the given form. …
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CBSE 2026Set CX1 mark
$\int x^{2}\tan(x^{3}+2)\,dx$
›Reveal solutionSolution

Substitute $u=x^{3}+2$ ; the integral becomes $\tfrac13\int\tan u\,du=\tfrac13\ln|\sec(x^{3}+2)|+C$ .

Concept: The factor $x^{2}$ is (up to a constant) the derivative of the inner function $x^{3}+2$ , so substitution works.

Let $u=x^{3}+2\Rightarrow du=3x^{2}\,dx\Rightarrow x^{2}\,dx=\dfrac{du}{3}$ .

…

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CBSE 2026Set A1 markMCQ
$\int\frac{\tan(\sin^{-1}x)}{\sqrt{1-x^2}}\,dx =$
(a) $\log|\sec(\sin^{-1}x)| + k$
(b) $\log|\cos(\sin^{-1}x)| + k$
(c) $\tan(\sin^{-1}x) + k$
(d) $\log|\sin^{-1}x| + k$
›Reveal solutionSolution

With $u=\sin^{-1}x$ (so $du=\tfrac{dx}{\sqrt{1-x^2}}$ ) the integral is $\int\tan u\,du=\log|\sec u|+k$ .

Let $u=\sin^{-1}x$ . Then $du=\dfrac{dx}{\sqrt{1-x^2}}$ , so

…

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CBSE 2026Set A1 markMCQ
$\int\frac{dx}{e^x + e^{-x}} =$
(a) $\cot^{-1}(e^x) + k$
(b) $\tan^{-1}(e^x) + k$
(c) $\log|e^x + 1| + k$
(d) $\sin^{-1}(e^x) + k$
›Reveal solutionSolution

Substitute $t=e^x$ : $\int\tfrac{dx}{e^x+e^{-x}}=\int\tfrac{dt}{1+t^2}=\tan^{-1}(e^x)+k$ .

Multiply numerator and denominator by $e^x$ :

$\dfrac{1}{e^x+e^{-x}}=\dfrac{e^x}{e^{2x}+1}$ .

Let $t=e^x$ , $dt=e^x\,dx$ . Then

…

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CBSE 2026Set A1 markMCQ
$\int_1^e\frac{(\log x)^2}{x}\,dx =$
(a) $\frac{1}{3}$
(b) $\frac{1}{3}e^3$
(c) $\frac{1}{3}(e^3 - 1)$
(d) $e^3$
›Reveal solutionSolution

Substitute $t=\log x$ : the integral becomes $\int_0^1 t^2\,dt=\tfrac13$ .

Let $t=\log x$ , so $dt=\tfrac{dx}{x}$ . Limits: $x=1\to t=0$ , $x=e\to t=1$ . Then

…

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CBSE 2026Set A1 markMCQ
$\int_0^{\pi/4}\frac{e^{\tan x}}{\cos^2 x}\,dx =$
(a) $e - 1$
(b) $e + 1$
(c) $\frac{1}{e} + 1$
(d) $\frac{1}{e} - 1$
›Reveal solutionSolution

Substitute $t=\tan x$ (so $dt=\sec^2 x\,dx$ ): the integral becomes $\int_0^1 e^t\,dt=e-1$ .

Let $t=\tan x$ . Then $dt=\sec^2 x\,dx=\dfrac{dx}{\cos^2 x}$ . Limits: $x=0\to t=0$ , $x=\tfrac{\pi}{4}\to t=1$ . So

…

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CBSE 2026Set A1 markMCQ
$\int_0^1\frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx =$
(a) $2(e - 1)$
(b) $e - 1$
(c) $2(e + 1)$
(d) $e + 1$
›Reveal solutionSolution

Substitute $t=\sqrt{x}$ ; the integral becomes $2\int_0^1 e^t\,dt=2(e-1)$ .

Let $t=\sqrt{x}$ , so $dt=\dfrac{dx}{2\sqrt{x}}$ , i.e. $\dfrac{dx}{\sqrt{x}}=2\,dt$ . Limits: $x=0\to t=0$ , $x=1\to t=1$ . Then

…

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CBSE 2026Set A1 markMCQ
$\int_0^a\sqrt{a^2 - x^2}\,dx =$
(a) $\frac{\pi}{4}$
(b) $\frac{a^2}{4}$
(c) $\frac{\pi a^2}{4}$
(d) $\pi$
›Reveal solutionSolution

$\int_0^a\sqrt{a^2-x^2}\,dx$ is a quarter-circle area $=\tfrac{\pi a^2}{4}$ .

Using the standard formula $\int\sqrt{a^2-x^2}\,dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}$ , evaluate from $0$ to $a$ :

…

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CBSE 2026Set ANNUAL1 markMCQ
$\int \sin(2x+3)\,dx =$
(a) $\cos(2x+3)+C$
(b) $-\dfrac{\cos(2x+3)}{2}+C$
(c) $\tan 2x + C$
(d) None of these
›Reveal solutionSolution

$\int \sin(ax+b)\,dx = -\dfrac{\cos(ax+b)}{a}+C$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\displaystyle\int_1^{e} \dfrac{(\log x)^2}{x}\, dx =$
(a) $\dfrac{1}{3}e^3$
(b) $\dfrac{1}{3}(e^3-1)$
(c) $\dfrac{1}{3}$
(d) None of these
›Reveal solutionSolution

Substitute $u = \log x$ , $du = dx/x$ , converting the limits from $x=1,e$ to $u=0,1$ .

Let $u=\log x \Rightarrow du = \dfrac{dx}{x}$ . When $x=1, u=0$ ; when $x=e, u=1$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{1}{x(1+\log x)}\,dx$
(a) $x + \log x + c$
(b) $|x + \log x| + c$
(c) $\log|1 + \log x| + c$
(d) $\log(1+x) + c$
›Reveal solutionSolution

Substitute $u=1+\log x$ so $du=\dfrac{dx}{x}$ , turning the integral into $\int\dfrac{du}{u}$ .

$I=\displaystyle\int\dfrac{1}{x(1+\log x)}\,dx$

Let $u=1+\log x \Rightarrow du=\dfrac{1}{x}dx$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{\sin^6 x}{\cos^8 x}\,dx$
›Reveal solutionSolution

Rewrite the integrand as $\tan^6x\sec^2x$ and substitute $t=\tan x$ .

$I=\displaystyle\int\dfrac{\sin^6x}{\cos^8x}\,dx=\int\dfrac{\sin^6x}{\cos^6x}\cdot\dfrac{1}{\cos^2x}\,dx=\int\tan^6x\sec^2x\,dx$

Let $t=\tan x \Rightarrow dt=\sec^2x\,dx$ . …

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Q.(a) Find : ∫cos⁡x(4+sin⁡2x)(5−4cos⁡2x) dx\int \frac{\cos x}{(4+\sin^2 x)(5-4\cos^2 x)}\, dx∫(4+sin2x)(5−4cos2x)cosx​dx

Part (a)

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Q.(a) Find : $\int \frac{\cos x}{(4+\sin^2 x)(5-4\cos^2 x)}\, dx$

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