If \int \sin x \over \sin ^3x + \cos ^3xdx = \alpha \log _e|1 + \tan x| + \beta \log _e|1 - \tan x + \tan ^2x| + \gamma \tan ^ - 1( 2\tan x - 1 \over \sqrt 3 ) + C , when C is constant of integration, then the value of 18(\alpha + \beta + \gamma ^2) is \underline\hspace2cm .

Indefinite Integrals PYQ — JEE Main 2021 · 31 Aug · Shift 2 · Q39 with Solution

View written solutionFree

Correct answer: 3

Given integral

We need to evaluate

$I=\int \frac{\sin x}{\sin^3 x+\cos^3 x}\,dx$

and compare it with

$I=\alpha \ln|1+\tan x|+\beta \ln|1-\tan x+\tan^2 x|+\gamma \tan^{-1}\!\left(\frac{2\tan x-1}{\sqrt3}\right)+C.$

Then compute

$18(\alpha+\beta+\gamma^2).$

Substitute $t=\tan x$

We use

$\sin x=t\cos x, \qquad dx=\frac{dt}{1+t^2}.$

Also,

$\sin^3x+\cos^3x=\cos^3x(t^3+1).$

Hence

=\frac{t\cos x}{\cos^3x(1+t^3)} =\frac{t}{\cos^2x(1+t^3)}.$$ Since $$\frac{1}{\cos^2x}=1+t^2,$$ we get $$\frac{\sin x}{\sin^3x+\cos^3x}dx =\frac{t(1+t^2)}{1+t^3}\cdot \frac{dt}{1+t^2} =\frac{t}{1+t^3}\,dt.$$ So $$I=\int \frac{t}{1+t^3}\,dt.$$ --- 3. **Factor the denominator** $$1+t^3=(t+1)(t^2-t+1).$$ So we decompose: $$\frac{t}{(t+1)(t^2-t+1)}=\frac{A}{t+1}+\frac{Bt+C}{t^2-t+1}.$$ Multiplying through, $$t=A(t^2-t+1)+(Bt+C)(t+1).$$ Expand: $$t=A t^2-A t+A + B t^2+B t + C t + C.$$ Thus $$t=(A+B)t^2+(-A+B+C)t+(A+C).$$ Comparing coefficients: $$A+B=0,$$ $$-A+B+C=1,$$ $$A+C=0.$$ From $A+C=0$, $C=-A$. From $A+B=0$, $B=-A$. Then $$-A+(-A)+(-A)=1 \implies -3A=1 \implies A=-\frac13.$$ Hence $$B=\frac13, \qquad C=\frac13.$$ Therefore, $$\frac{t}{1+t^3}=-\frac{1}{3(t+1)}+\frac{t+1}{3(t^2-t+1)}.$$ So $$I=-\frac13\int \frac{dt}{t+1}+\frac13\int \frac{t+1}{t^2-t+1}\,dt.$$ --- 4. **Integrate the second term** Write $$t+1=\frac12(2t-1)+\frac32.$$ Therefore, $$\int \frac{t+1}{t^2-t+1}\,dt =\frac12\int \frac{2t-1}{t^2-t+1}\,dt+\frac32\int \frac{dt}{t^2-t+1}.$$ Now, $$\int \frac{2t-1}{t^2-t+1}\,dt=\ln|t^2-t+1|.$$ Also, $$t^2-t+1=\left(t-\frac12\right)^2+\frac34 =\frac{(2t-1)^2+3}{4}.$$ Hence $$\int \frac{dt}{t^2-t+1} =\int \frac{dt}{\left(t-\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2} =\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2t-1}{\sqrt3}\right).$$ So, $$\int \frac{t+1}{t^2-t+1}\,dt =\frac12\ln|t^2-t+1|+\sqrt3\tan^{-1}\left(\frac{2t-1}{\sqrt3}\right).$$ Multiply by $\frac13$: $$\frac13\int \frac{t+1}{t^2-t+1}\,dt =\frac16\ln|t^2-t+1|+\frac{\sqrt3}{3}\tan^{-1}\left(\frac{2t-1}{\sqrt3}\right).$$ Thus $$I=-\frac13\ln|t+1|+\frac16\ln|t^2-t+1|+\frac{\sqrt3}{3}\tan^{-1}\left(\frac{2t-1}{\sqrt3}\right)+C.$$ Substitute back $t=\tan x$: $$I=-\frac13\ln|1+\tan x|+\frac16\ln|1-\tan x+\tan^2x|+\frac{1}{\sqrt3}\tan^{-1}\left(\frac{2\tan x-1}{\sqrt3}\right)+C.$$ So, $$\alpha=-\frac13,\qquad \beta=\frac16,\qquad \gamma=\frac{1}{\sqrt3}.$$ --- 5. **Compute the required value** $$\gamma^2=\frac13.$$ Then $$\alpha+\beta+\gamma^2=-\frac13+\frac16+\frac13=\frac16.$$ Therefore, $$18(\alpha+\beta+\gamma^2)=18\cdot \frac16=3.$$ --- 6. **Comparison with stored answer** Our derived answer is $3$, which matches the stored correct answer.

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View written solutionFree

Correct answer: 3

Given integral

We need to evaluate

$I=\int \frac{\sin x}{\sin^3 x+\cos^3 x}\,dx$

and compare it with

$I=\alpha \ln|1+\tan x|+\beta \ln|1-\tan x+\tan^2 x|+\gamma \tan^{-1}\!\left(\frac{2\tan x-1}{\sqrt3}\right)+C.$

Then compute

$18(\alpha+\beta+\gamma^2).$

Substitute $t=\tan x$

We use

$\sin x=t\cos x, \qquad dx=\frac{dt}{1+t^2}.$

Also,

$\sin^3x+\cos^3x=\cos^3x(t^3+1).$

Hence

=\frac{t\cos x}{\cos^3x(1+t^3)} =\frac{t}{\cos^2x(1+t^3)}.$$ Since $$\frac{1}{\cos^2x}=1+t^2,$$ we get $$\frac{\sin x}{\sin^3x+\cos^3x}dx =\frac{t(1+t^2)}{1+t^3}\cdot \frac{dt}{1+t^2} =\frac{t}{1+t^3}\,dt.$$ So $$I=\int \frac{t}{1+t^3}\,dt.$$ --- 3. **Factor the denominator** $$1+t^3=(t+1)(t^2-t+1).$$ So we decompose: $$\frac{t}{(t+1)(t^2-t+1)}=\frac{A}{t+1}+\frac{Bt+C}{t^2-t+1}.$$ Multiplying through, $$t=A(t^2-t+1)+(Bt+C)(t+1).$$ Expand: $$t=A t^2-A t+A + B t^2+B t + C t + C.$$ Thus $$t=(A+B)t^2+(-A+B+C)t+(A+C).$$ Comparing coefficients: $$A+B=0,$$ $$-A+B+C=1,$$ $$A+C=0.$$ From $A+C=0$, $C=-A$. From $A+B=0$, $B=-A$. Then $$-A+(-A)+(-A)=1 \implies -3A=1 \implies A=-\frac13.$$ Hence $$B=\frac13, \qquad C=\frac13.$$ Therefore, $$\frac{t}{1+t^3}=-\frac{1}{3(t+1)}+\frac{t+1}{3(t^2-t+1)}.$$ So $$I=-\frac13\int \frac{dt}{t+1}+\frac13\int \frac{t+1}{t^2-t+1}\,dt.$$ --- 4. **Integrate the second term** Write $$t+1=\frac12(2t-1)+\frac32.$$ Therefore, $$\int \frac{t+1}{t^2-t+1}\,dt =\frac12\int \frac{2t-1}{t^2-t+1}\,dt+\frac32\int \frac{dt}{t^2-t+1}.$$ Now, $$\int \frac{2t-1}{t^2-t+1}\,dt=\ln|t^2-t+1|.$$ Also, $$t^2-t+1=\left(t-\frac12\right)^2+\frac34 =\frac{(2t-1)^2+3}{4}.$$ Hence $$\int \frac{dt}{t^2-t+1} =\int \frac{dt}{\left(t-\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2} =\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2t-1}{\sqrt3}\right).$$ So, $$\int \frac{t+1}{t^2-t+1}\,dt =\frac12\ln|t^2-t+1|+\sqrt3\tan^{-1}\left(\frac{2t-1}{\sqrt3}\right).$$ Multiply by $\frac13$: $$\frac13\int \frac{t+1}{t^2-t+1}\,dt =\frac16\ln|t^2-t+1|+\frac{\sqrt3}{3}\tan^{-1}\left(\frac{2t-1}{\sqrt3}\right).$$ Thus $$I=-\frac13\ln|t+1|+\frac16\ln|t^2-t+1|+\frac{\sqrt3}{3}\tan^{-1}\left(\frac{2t-1}{\sqrt3}\right)+C.$$ Substitute back $t=\tan x$: $$I=-\frac13\ln|1+\tan x|+\frac16\ln|1-\tan x+\tan^2x|+\frac{1}{\sqrt3}\tan^{-1}\left(\frac{2\tan x-1}{\sqrt3}\right)+C.$$ So, $$\alpha=-\frac13,\qquad \beta=\frac16,\qquad \gamma=\frac{1}{\sqrt3}.$$ --- 5. **Compute the required value** $$\gamma^2=\frac13.$$ Then $$\alpha+\beta+\gamma^2=-\frac13+\frac16+\frac13=\frac16.$$ Therefore, $$18(\alpha+\beta+\gamma^2)=18\cdot \frac16=3.$$ --- 6. **Comparison with stored answer** Our derived answer is $3$, which matches the stored correct answer.

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