Q.Integrate the following function: $\sin^3 x \cos^3 x$

Showing the 12 most recent of 22 on this concept.

1. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.$\displaystyle\int\sin^4x\cos^4x\,dx=$ (A) $\dfrac{1}{128}(-2\sin^3x\cos x-3\sin x\cos x+3)+c$ (B) $\dfrac{1}{256}(-2\sin^3 2x\cos2x-3\sin2x\cos2x+6x)+c$ (C) $\dfrac{1}{128}(2\sin^3x\cos x-3\sin x\cos x+3x)+c$ (D) $\dfrac{1}{256}(3\sin^3x\cos x-2\sin x\cos x+2)+c$

   ›Reveal solutionSolution

   This tests reduction of $\sin^4x\cos^4x$ via the double-angle substitution $\sin x\cos x=\frac12\sin2x$; the answer is (B).

   Concept and Intuition

   Products of even powers of $\sin x,\cos x$ are attacked by writing $\sin x\cos x=\tfrac12\sin2x$ first, which collapses the problem to a single power of $\sin(2x)$, and then a standard power-reduction formula finishes it.

   Step-by-Step Solution

   1. $\sin^4x\cos^4x=(\sin x\cos x)^4=\left(\frac{\sin2x}{2}\right)^4=\frac{1}{16}\sin^4(2x)$.
   2. Use $\sin^4u=\dfrac{3}{8}-\dfrac12\cos2u+\dfrac18\cos4u$ with $u=2x$: $\sin^4(2x)=\dfrac38-\dfrac12\cos4x+\dfrac18\cos8x$.
   3. So the integrand is $\dfrac{1}{16}\left(\dfrac38-\dfrac12\cos4x+\dfrac18\cos8x\right)$. Integrating term by term:

   $\int\sin^4x\cos^4x\,dx=\frac{1}{16}\left(\frac{3x}{8}-\frac{\sin4x}{8}+\frac{\sin8x}{64}\right)+c=\frac{3x}{128}-\frac{\sin4x}{128}+\frac{\sin8x}{1024}+c.$

   4. Convert to functions of $2x$: $\sin4x=2\sin2x\cos2x$, and $\sin8x=2\sin4x\cos4x=4\sin2x\cos2x(1-2\sin^2 2x)=4\sin2x\cos2x-8\sin^3(2x)\cos2x$. …

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2. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\int \frac{1+x^2}{\sqrt{1-x^2}}dx =$ (A) $\frac{3}{2}\mathrm{Sin}^{-1}x - \frac{x}{2}\sqrt{1-x^2} + c$ (B) $\frac{3}{2}\mathrm{Sin}^{-1}x + \frac{x}{2}\sqrt{1-x^2} + c$ (C) $\frac{3}{4}\mathrm{Sin}^{-1}x - \frac{x^2}{2}\sqrt{1-x^2} + c$ (D) $\frac{2}{3}\mathrm{Sin}^{-1}x + \frac{x^2}{2}\sqrt{1-x^2} + c$

   ›Reveal solutionSolution

   Rewriting $1+x^2$ as $2-(1-x^2)$ turns an awkward integral into a difference of two standard forms, giving $\frac32\sin^{-1}x-\frac{x}{2}\sqrt{1-x^2}+c$.

   Concept and Intuition

   The integrand mixes $1$ and $x^2$ with a $\sqrt{1-x^2}$ denominator. The trick is to notice that $1-x^2$ itself appears under the root, so rewriting the numerator in terms of $(1-x^2)$ converts the whole integral into two textbook-standard integrals: $\int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x$ and $\int\sqrt{1-x^2}\,dx=\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x$.

   Step-by-Step Solution

   1. Write $1+x^2 = 2-(1-x^2)$.
   2. So $\displaystyle\int\frac{1+x^2}{\sqrt{1-x^2}}dx=\int\frac{2}{\sqrt{1-x^2}}dx-\int\frac{1-x^2}{\sqrt{1-x^2}}dx = 2\int\frac{dx}{\sqrt{1-x^2}}-\int\sqrt{1-x^2}\,dx$.
   3. Use the standard results: $\int\dfrac{dx}{\sqrt{1-x^2}}=\sin^{-1}x+c_1$ and $\int\sqrt{1-x^2}\,dx=\dfrac{x}{2}\sqrt{1-x^2}+\dfrac12\sin^{-1}x+c_2$. …

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3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{\cos 2x + \sin 2x + 2\sin^2 x} =$ (A) $-\dfrac{1}{1 + \tan x} + c$ (B) $-\dfrac{\tan x}{1 + \tan x} + c$ (C) $-\cot x + c$ (D) $-\tan x + c$

   ›Reveal solutionSolution

   The key trick is recognising that $\cos2x+2\sin^2x\equiv1$, collapsing the denominator to $1+\sin2x=(\sin x+\cos x)^2$, after which a $\tan x$ substitution finishes it. Answer: $-\dfrac{1}{1+\tan x}+c$.

   Concept and Intuition

   Many trig-integral MCQs hide a simplification of the denominator using a double-angle identity. Spotting $\cos2x=1-2\sin^2x$ instantly cancels the $2\sin^2x$ term here, and then $1+\sin2x$ is a perfect square $(\sin x+\cos x)^2$ — a very standard identity worth memorising.

   Step-by-Step Solution

   1. $\cos2x+\sin2x+2\sin^2x=(\cos2x+2\sin^2x)+\sin2x=1+\sin2x$ (using $\cos2x=1-2\sin^2x$).
   2. $1+\sin2x=1+2\sin x\cos x=(\sin x+\cos x)^2$. …

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4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.$\displaystyle\int_{\pi/4}^{\pi/3} \dfrac{\cos x-\sin x}{\sin 2x}\,dx =$ (A) $\dfrac{1}{2}\log\left[\dfrac{(3+2\sqrt{2})(2-\sqrt{3})}{\sqrt{3}}\right]$ (B) $\dfrac{1}{2}\log\left[\dfrac{(3-2\sqrt{2})(2+\sqrt{3})}{\sqrt{3}}\right]$ (C) $\log\left[\dfrac{(3-2\sqrt{2})(2-\sqrt{3})}{\sqrt{3}}\right]$ (D) $\log\left[\dfrac{(3+2\sqrt{2})(2-\sqrt{3})}{\sqrt{3}}\right]$

   ›Reveal solutionSolution

   This tests splitting a trig rational integrand using $\sin 2x=2\sin x\cos x$ into standard $\csc$/$\sec$ integrals; the answer is (A).

   Concept and Intuition

   Whenever you see $\sin 2x$ in a denominator paired with $\cos x - \sin x$ in the numerator, it is almost always meant to be split via the double-angle identity into two elementary reciprocal-trig integrals, each of which has a well-known logarithmic antiderivative.

   Step-by-Step Solution

   1. Write $\sin 2x = 2\sin x \cos x$, so

   $\frac{\cos x-\sin x}{\sin 2x}=\frac{\cos x-\sin x}{2\sin x\cos x}=\frac12\left(\frac{1}{\sin x}-\frac{1}{\cos x}\right)=\frac12(\csc x-\sec x).$

   2. Use the standard antiderivatives $\int \csc x\,dx=-\log|\csc x+\cot x|$ and $\int\sec x\,dx=\log|\sec x+\tan x|$. So

   $\int\frac12(\csc x-\sec x)\,dx=-\frac12\log\big[(\csc x+\cot x)(\sec x+\tan x)\big]+C.$

   3. At $x=\pi/3$: $\csc x+\cot x=\frac{2}{\sqrt3}+\frac1{\sqrt3}=\sqrt3$, and $\sec x+\tan x=2+\sqrt3$. Product $=\sqrt3(2+\sqrt3)=2\sqrt3+3$.
   4. At $x=\pi/4$: $\csc x+\cot x=\sqrt2+1$, and $\sec x+\tan x=\sqrt2+1$. Product $=(\sqrt2+1)^2=3+2\sqrt2$.
   5. So the definite integral $=-\frac12\left[\log(2\sqrt3+3)-\log(3+2\sqrt2)\right]=\frac12\log\dfrac{3+2\sqrt2}{2\sqrt3+3}$. …

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5. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin 2x}{\sin^2 x+3\cos x-3}\,dx =$ (A) $2\log\left|\dfrac{\cos x-2}{\cos x-1}\right|+c$ (B) $\log\left(\dfrac{(\cos x-2)^2}{(\cos x-1)^4}\right)+c$ (C) $\log\left(\dfrac{(\cos x-2)^2}{|\cos x-1|}\right)+c$ (D) $\log\left(\dfrac{(\cos x-2)^4}{(\cos x-1)^2}\right)+c$

   ›Reveal solutionSolution

   Factor the denominator using $\sin^2x=1-\cos^2x$, substitute $u=\cos x$, and resolve the rational integrand by partial fractions. Answer: $\log\!\left(\dfrac{(\cos x-2)^4}{(\cos x-1)^2}\right)+c$.

   Concept and Intuition

   The presence of $\sin2x\,dx=2\sin x\cos x\,dx$ alongside a denominator built purely from $\cos x$ (after using the Pythagorean identity) signals the substitution $u=\cos x$: then $du=-\sin x\,dx$ turns the whole thing into a rational function of $u$, solvable by partial fractions.

   Step-by-Step Solution

   1. Rewrite the denominator: $\sin^2x+3\cos x-3 = (1-\cos^2x)+3\cos x-3 = -\cos^2x+3\cos x-2 = -(\cos^2x-3\cos x+2)$.
   2. Factor the quadratic: $\cos^2x-3\cos x+2=(\cos x-1)(\cos x-2)$. So the denominator is $-(\cos x-1)(\cos x-2)$.
   3. Let $u=\cos x \Rightarrow du = -\sin x\,dx$. Also $\sin2x = 2\sin x\cos x = 2u\sin x$.
   4. Rewrite the integral:

   $\int \frac{2u\sin x\,dx}{-(u-1)(u-2)}.$

   Since $du=-\sin x\,dx \Rightarrow \sin x\,dx = -du$:

   $\int \frac{2u\cdot(-du)}{-(u-1)(u-2)} = \int \frac{2u\,du}{(u-1)(u-2)}.$

   5. Partial fractions: $\dfrac{2u}{(u-1)(u-2)}=\dfrac{A}{u-1}+\dfrac{B}{u-2}$. Then $2u=A(u-2)+B(u-1)$. At $u=1$: $2=-A\Rightarrow A=-2$. At $u=2$: $4=B\Rightarrow B=4$.
   6. Integrate: $\displaystyle\int\left(\frac{-2}{u-1}+\frac{4}{u-2}\right)du = -2\log|u-1|+4\log|u-2|+c$. …

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6. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.$\displaystyle\int_{0}^{\pi} \left(\cos^2\left(\dfrac{3\pi}{8} - \dfrac{x}{4}\right) - \cos^2\left(\dfrac{11\pi}{8} + \dfrac{x}{4}\right)\right) dx =$ (A) $1/\sqrt{2}$ (B) $2\sqrt{2}$ (C) $\sqrt{2}$ (D) $2$

   ›Reveal solutionSolution

   The difference of squared cosines collapses via a product-to-sum identity to $\tfrac{\sqrt2}{2}\sin(x/2)$, which integrates to $\sqrt2$ over $[0,\pi]$.

   Concept and Intuition

   $\cos^2A-\cos^2B$ always simplifies to $-\sin(A+B)\sin(A-B)$ (from the double-angle formula $\cos^2\theta=\tfrac{1+\cos2\theta}{2}$ and the sum-to-product identity for cosines). Applying this collapses the two ugly-looking angle expressions into a single simple sine of $x/2$.

   Step-by-Step Solution

   1. Let $A=\dfrac{3\pi}{8}-\dfrac{x}{4}$, $B=\dfrac{11\pi}{8}+\dfrac{x}{4}$.
   2. $\cos^2A-\cos^2B = \dfrac{1+\cos2A}{2}-\dfrac{1+\cos2B}{2} = \dfrac{\cos2A-\cos2B}{2} = -\sin(A+B)\sin(A-B)$ (using $\cos P-\cos Q = -2\sin\tfrac{P+Q}{2}\sin\tfrac{P-Q}{2}$ with $P=2A,Q=2B$).
   3. $A+B = \dfrac{3\pi}{8}+\dfrac{11\pi}{8} = \dfrac{14\pi}{8}=\dfrac{7\pi}{4}$; the $x$ terms cancel.
   4. $A-B = \dfrac{3\pi}{8}-\dfrac{11\pi}{8} - \dfrac{x}{4}-\dfrac{x}{4} = -\pi - \dfrac{x}{2}$.
   5. So integrand $= -\sin\!\left(\dfrac{7\pi}{4}\right)\sin\!\left(-\pi-\dfrac{x}{2}\right)$. Now $\sin(7\pi/4)=-\dfrac{\sqrt2}{2}$ and $\sin(-\pi-x/2) = -\sin(\pi+x/2) = -(-\sin(x/2)) = \sin(x/2)$. …

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7. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{\sin 6x}{\sin x}\,dx =$ (A) $\dfrac{26}{15}$ (B) $\dfrac{256}{15}$ (C) $\dfrac{64}{15}$ (D) $\dfrac{17}{15}$

   ›Reveal solutionSolution

   Expanding $\sin6x/\sin x$ as a sum of cosines (a Chebyshev-type identity) and integrating term by term gives $26/15$.

   Concept and Intuition

   Ratios like $\sin(n\theta)/\sin\theta$ are polynomials in $\cos\theta$ (Chebyshev polynomials of the second kind), and can always be rewritten as a finite sum of cosines of multiples of $\theta$. This makes the integral trivial term by term, instead of trying to integrate the ratio directly.

   Step-by-Step Solution

   1. Claim: $\sin6\theta=\sin\theta\,[2\cos5\theta+2\cos3\theta+2\cos\theta]$. Verify using the product-to-sum identity $2\sin\theta\cos k\theta=\sin(k+1)\theta-\sin(k-1)\theta$: taking $k=5,3,1$ gives $(\sin6\theta-\sin4\theta)+(\sin4\theta-\sin2\theta)+(\sin2\theta-\sin0)=\sin6\theta$ — the sum telescopes perfectly.
   2. So $\dfrac{\sin6x}{\sin x}=2\cos5x+2\cos3x+2\cos x$.
   3. Integrate: $\displaystyle\int_0^{\pi/2}2\cos5x\,dx=\left[\tfrac{2}{5}\sin5x\right]_0^{\pi/2}=\tfrac25\sin\tfrac{5\pi}{2}=\tfrac25(1)=\tfrac25$. …

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8. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.$\int_{-\pi}^{\pi/2}\sin x\cdot\sin^2(\cos x)\,dx=$ (A) $\dfrac{1-\sin2}{4}$ (B) $-\left(\dfrac{1+\sin2}{4}\right)$ (C) $\dfrac{\sin2-2}{4}$ (D) $-\left(\dfrac{2+\sin2}{4}\right)$

   ›Reveal solutionSolution

   A substitution $u=\cos x$ turns $\sin x\sin^2(\cos x)$ into a plain $\sin^2u$ integral over $[-1,0]$; the value works out to $\dfrac{\sin2-2}{4}$.

   Concept and Intuition

   Whenever an integrand contains $\sin x\cdot f(\cos x)$, the derivative of $\cos x$ (which is $-\sin x$) is already sitting there — that's the signal to substitute $u=\cos x$. It converts a trig-of-trig integral into an ordinary polynomial-in-$\sin u$ integral.

   Step-by-Step Solution

   1. Let $u=\cos x\Rightarrow du=-\sin x\,dx$, i.e. $\sin x\,dx=-du$.
   2. Limits: at $x=-\pi$, $u=\cos(-\pi)=-1$; at $x=\pi/2$, $u=\cos(\pi/2)=0$.
   3. $\displaystyle I=\int_{-\pi}^{\pi/2}\sin x\sin^2(\cos x)\,dx=\int_{u=-1}^{0}\sin^2u\cdot(-du)=-\int_{-1}^{0}\sin^2u\,du$.
   4. Use $\sin^2u=\dfrac{1-\cos2u}{2}$, so $\displaystyle\int\sin^2u\,du=\frac{u}{2}-\frac{\sin2u}{4}$.
   5. Evaluate: at $u=0$ this is $0$; at $u=-1$ this is $-\dfrac12-\dfrac{\sin(-2)}{4}=-\dfrac12+\dfrac{\sin2}{4}$. …

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9. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin\alpha}{\sqrt{1+\cos\alpha}}\, d\alpha =$ (A) $-2\sqrt{2}\cos\left(\dfrac{\alpha}{2}\right) + c$ (B) $2\sqrt{2}\cos\left(\dfrac{\alpha}{2}\right) + c$ (C) $\sqrt{2}\cos\left(\dfrac{\alpha}{2}\right) + c$ (D) $-\sqrt{2}\cos\left(\dfrac{\alpha}{2}\right) + c$

   ›Reveal solutionSolution

   This tests a simple substitution followed by the half-angle identity $1+\cos\alpha = 2\cos^2(\alpha/2)$. Answer: $-2\sqrt2\cos(\alpha/2)+c$.

   Concept and Intuition

   When the integrand contains $\sin\alpha$ alongside a function of $\cos\alpha$, substituting $u=1+\cos\alpha$ turns it into an elementary power-rule integral; the half-angle identity then lets us re-express the result in the compact closed form matching the options.

   Step-by-Step Solution

   1. Let $u = 1+\cos\alpha$, so $du = -\sin\alpha\,d\alpha$, i.e. $\sin\alpha\,d\alpha = -du$.
   2. The integral becomes $\displaystyle\int \dfrac{-du}{\sqrt u} = -\int u^{-1/2}\,du = -2u^{1/2}+c = -2\sqrt{1+\cos\alpha}+c$.
   3. Use the half-angle identity: $1+\cos\alpha = 2\cos^2\left(\dfrac{\alpha}{2}\right)$, so $\sqrt{1+\cos\alpha} = \sqrt2\left|\cos\left(\dfrac{\alpha}{2}\right)\right|$. …

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10. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.If $\int\dfrac{\sin^2\alpha-\sin^2x}{\cos x-\cos\alpha}dx=f(x)+Ax+B$ and $B\in\mathbb{R}$, then (A) $f(x)=2\sin x,\ A=\cos\alpha$ (B) $f(x)=2\sin x,\ A=2\cos\alpha$ (C) $f(x)=\sin x,\ A=\cos\alpha$ (D) $f(x)=\sin x,\ A=2\cos\alpha$

    ›Reveal solutionSolution

    The fraction $\dfrac{\sin^2\alpha-\sin^2x}{\cos x-\cos\alpha}$ simplifies via standard trig identities to the constant-plus-cosine expression $\cos x+\cos\alpha$, which integrates trivially.

    Concept and Intuition

    Two very standard identities do all the work: $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$ and $\cos B-\cos A=2\sin\frac{A+B}2\sin\frac{A-B}2$. Expressing both the numerator and denominator this way lets most factors cancel, leaving a simple product-to-sum term.

    Step-by-Step Solution

    1. $\sin^2\alpha-\sin^2x=\sin(\alpha+x)\sin(\alpha-x)$.
    2. Write $\sin(\alpha+x)=2\sin\frac{\alpha+x}2\cos\frac{\alpha+x}2$ and $\sin(\alpha-x)=2\sin\frac{\alpha-x}2\cos\frac{\alpha-x}2$, so the numerator becomes $4\sin\frac{\alpha+x}2\cos\frac{\alpha+x}2\sin\frac{\alpha-x}2\cos\frac{\alpha-x}2$.
    3. $\cos x-\cos\alpha=-2\sin\frac{x+\alpha}2\sin\frac{x-\alpha}2=2\sin\frac{\alpha+x}2\sin\frac{\alpha-x}2$ (flipping the sign of the second sine turns $\sin\frac{x-\alpha}2=-\sin\frac{\alpha-x}2$).
    4. Dividing: $\dfrac{4\sin\frac{\alpha+x}2\cos\frac{\alpha+x}2\sin\frac{\alpha-x}2\cos\frac{\alpha-x}2}{2\sin\frac{\alpha+x}2\sin\frac{\alpha-x}2}=2\cos\frac{\alpha+x}2\cos\frac{\alpha-x}2$. …

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11. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
    Q.If $\int \frac{\cos 8x + 1}{\cot 2x - \tan 2x}dx = A\cos 8x + c$, then $A =$ (A) $-\frac{1}{16}$ (B) $\frac{1}{16}$ (C) $-\frac{1}{8}$ (D) $\frac{1}{8}$

    ›Reveal solutionSolution

    Simplifying the trigonometric expression using double-angle identities collapses the integrand to $\tfrac12\sin8x$, whose antiderivative directly gives $A=-\tfrac1{16}$.

    Concept and Intuition

    This problem looks intimidating because of the $\cot2x-\tan2x$ combination, but that combination has a clean double-angle identity: $\cot\theta-\tan\theta=2\cot2\theta$. Combined with $1+\cos8x=2\cos^24x$, nearly everything cancels, leaving a single simple sine term to integrate.

    Step-by-Step Solution

    1. Simplify the denominator: $\cot2x-\tan2x=\dfrac{\cos2x}{\sin2x}-\dfrac{\sin2x}{\cos2x}=\dfrac{\cos^22x-\sin^22x}{\sin2x\cos2x}=\dfrac{\cos4x}{\tfrac12\sin4x}=2\cot4x$.
    2. Simplify the numerator: $\cos8x+1=2\cos^24x$ (using $\cos2\theta=2\cos^2\theta-1$ with $\theta=4x$).
    3. So the integrand is $\dfrac{2\cos^24x}{2\cot4x}=\cos^24x\cdot\tan4x=\cos4x\sin4x$.
    4. Use $\sin\theta\cos\theta=\tfrac12\sin2\theta$: $\cos4x\sin4x=\tfrac12\sin8x$. …

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12. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
    Q.If $\displaystyle\int \dfrac{dx}{2\sin^2 x+\sin 2x} = \dfrac{1}{2}\log|f(x)|+c$ and $f\left(\dfrac{\pi}{4}\right)=\dfrac{1}{2}$, then $f(x)=$ (A) $\dfrac{\sin x}{1+\sin x}$ (B) $\dfrac{\cos x}{1+\cos x}$ (C) $\dfrac{\tan x}{1+\tan x}$ (D) $\dfrac{\cot x}{1+\cot x}$

    ›Reveal solutionSolution

    Dividing the denominator by $\cos^2x$ turns this into a rational integral in $t=\tan x$; partial fractions give $f(x)=\dfrac{\tan x}{1+\tan x}$, matching the given boundary condition at $x=\pi/4$.

    Concept and Intuition

    $2\sin^2x+\sin2x$ factors as $2\sin x(\sin x+\cos x)$. Multiplying through by $\sec^2x/\sec^2x$ converts everything into functions of $\tan x$ alone, since $\sin x(\sin x+\cos x)\sec^2x = \tan^2x+\tan x$ — a purely algebraic (polynomial) expression in $t=\tan x$, letting us integrate as a simple rational function.

    Step-by-Step Solution

    1. Write $2\sin^2x+\sin2x = 2\sin^2x+2\sin x\cos x = 2\sin x(\sin x+\cos x)$.
    2. Multiply numerator and denominator by $\sec^2x$: since $\sin x(\sin x+\cos x)\sec^2x=\dfrac{\sin^2x}{\cos^2x}+\dfrac{\sin x\cos x}{\cos^2x}=\tan^2x+\tan x$, we get

    $\int\frac{dx}{2\sin x(\sin x+\cos x)}=\int\frac{\sec^2x\,dx}{2(\tan^2x+\tan x)}.$

    3. Let $t=\tan x$, $dt=\sec^2x\,dx$: the integral becomes $\dfrac12\displaystyle\int\dfrac{dt}{t(t+1)}$.
    4. Partial fractions: $\dfrac{1}{t(t+1)}=\dfrac1t-\dfrac{1}{t+1}$. …

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