The integral \int\limits_0^\pi / 4 \frac136 \sin x3 \sin x+5 \cos x ~d x is equal to :

3 \pi-25 \log _e 2+10 \log _e 5

3 \pi-10 \log _e(2 \sqrt2)+10 \log _e 5

3 \pi-30 \log _e 2+20 \log _e 5

Definite Integration PYQ — JEE Main 2024 · 5 Apr · Shift 1 · Q41 with Solution

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Correct answer: A

We need to evaluate $I=\int_0^{\pi/4} \frac{136\sin x}{3\sin x+5\cos x}\,dx.$
Rewrite the numerator in terms of the denominator.

We want constants $A,B$ such that $136\sin x=A(3\sin x+5\cos x)+B\cos x.$ Comparing coefficients:

for $\sin x$ : $3A=136 \Rightarrow A=\frac{136}{3}$ .
for $\cos x$ : $5A+B=0 \Rightarrow B=-5A=-\frac{680}{3}$ .

So, $\frac{136\sin x}{3\sin x+5\cos x}=\frac{136}{3}-\frac{680}{3}\cdot\frac{\cos x}{3\sin x+5\cos x}.$ Hence $I=\frac{136}{3}\int_0^{\pi/4}dx-\frac{680}{3}\int_0^{\pi/4}\frac{\cos x}{3\sin x+5\cos x}\,dx.$ Thus $I=\frac{136}{3}\cdot\frac{\pi}{4}-\frac{680}{3}J=\frac{34\pi}{3}-\frac{680}{3}J,$ where $J=\int_0^{\pi/4}\frac{\cos x}{3\sin x+5\cos x}\,dx.$

Evaluate $J$ .

Let $t=3\sin x+5\cos x.$ Then $dt=(3\cos x-5\sin x)dx,$ which does not directly match. So instead, use the tangent substitution: $u=\tan x,\quad dx=\frac{du}{1+u^2},\quad \sin x=\frac{u}{\sqrt{1+u^2}},\quad \cos x=\frac{1}{\sqrt{1+u^2}}.$ Then $3\sin x+5\cos x=\frac{3u+5}{\sqrt{1+u^2}}.$ Therefore,

=\frac{\frac1{\sqrt{1+u^2}}}{\frac{3u+5}{\sqrt{1+u^2}}}\cdot \frac{du}{1+u^2} =\frac{1}{(3u+5)(1+u^2)}\,du.$$ The limits change as: - $x=0 \Rightarrow u=0$, - $x=\pi/4 \Rightarrow u=1$. So $$J=\int_0^1 \frac{du}{(3u+5)(1+u^2)}.$$ 4. Partial fractions. Write $$\frac{1}{(3u+5)(1+u^2)}=\frac{Au+B}{1+u^2}+\frac{C}{3u+5}.$$ Then $$1=(Au+B)(3u+5)+C(1+u^2).$$ Expanding, $$1=(3A+C)u^2+(5A+3B)u+(5B+C).$$ Comparing coefficients: $$3A+C=0,$$ $$5A+3B=0,$$ $$5B+C=1.$$ From $C=-3A$ and $3B=-5A\Rightarrow B=-\frac{5A}{3}$. Substitute into $5B+C=1$: $$5\left(-\frac{5A}{3}\right)-3A=1 \Rightarrow -\frac{25A}{3}-3A=1 \Rightarrow -\frac{34A}{3}=1 \Rightarrow A=-\frac{3}{34}.$$ Thus $$B=\frac{5}{34},\qquad C=\frac{9}{34}.$$ So $$J=\int_0^1\left(\frac{-\frac{3}{34}u+\frac{5}{34}}{1+u^2}+\frac{9/34}{3u+5}\right)du.$$ That is, $$J=\frac{1}{34}\int_0^1\frac{-3u+5}{1+u^2}\,du+\frac{9}{34}\int_0^1\frac{du}{3u+5}.$$ 5. Integrate termwise. First, $$\int \frac{-3u+5}{1+u^2}du=-\frac{3}{2}\ln(1+u^2)+5\tan^{-1}u.$$ Next, $$\int \frac{du}{3u+5}=\frac13\ln(3u+5).$$ Therefore, $$J=\frac{1}{34}\left[-\frac{3}{2}\ln(1+u^2)+5\tan^{-1}u\right]_0^1+ rac{9}{34}\cdot\frac13\left[\ln(3u+5)\right]_0^1.$$ So $$J=\frac{1}{34}\left(-\frac{3}{2}\ln 2+5\cdot\frac{\pi}{4}\right)+\frac{3}{34}(\ln 8-\ln 5).$$ Since $\ln 8=3\ln 2$, $$J=\frac{5\pi}{136}-\frac{3\ln 2}{68}+\frac{9\ln 2}{34}-\frac{3\ln 5}{34}.$$ Now combine the $\ln 2$ terms: $$-\frac{3}{68}+\frac{18}{68}=\frac{15}{68}.$$ Hence $$J=\frac{5\pi}{136}+\frac{15}{68}\ln 2-\frac{3}{34}\ln 5.$$ 6. Substitute into $I$. $$I=\frac{34\pi}{3}-\frac{680}{3}\left(\frac{5\pi}{136}+\frac{15}{68}\ln 2-\frac{3}{34}\ln 5\right).$$ Now simplify each term: $$\frac{680}{3}\cdot \frac{5\pi}{136}=\frac{25\pi}{3},$$ $$\frac{680}{3}\cdot \frac{15}{68}=50,$$ $$\frac{680}{3}\cdot \frac{3}{34}=20.$$ Thus $$I=\frac{34\pi}{3}-\frac{25\pi}{3}-50\ln 2+20\ln 5.$$ So $$I=3\pi-50\ln 2+20\ln 5.$$ 7. Compare with options. This matches: $$\boxed{\text{A: }3\pi-50\log_e 2+20\log_e 5}.$$

More from Definite Integration

View written solutionFree

Correct answer: A

We need to evaluate $I=\int_0^{\pi/4} \frac{136\sin x}{3\sin x+5\cos x}\,dx.$
Rewrite the numerator in terms of the denominator.

We want constants $A,B$ such that $136\sin x=A(3\sin x+5\cos x)+B\cos x.$ Comparing coefficients:

for $\sin x$ : $3A=136 \Rightarrow A=\frac{136}{3}$ .
for $\cos x$ : $5A+B=0 \Rightarrow B=-5A=-\frac{680}{3}$ .

Evaluate $J$ .

=\frac{\frac1{\sqrt{1+u^2}}}{\frac{3u+5}{\sqrt{1+u^2}}}\cdot \frac{du}{1+u^2} =\frac{1}{(3u+5)(1+u^2)}\,du.$$ The limits change as: - $x=0 \Rightarrow u=0$, - $x=\pi/4 \Rightarrow u=1$. So $$J=\int_0^1 \frac{du}{(3u+5)(1+u^2)}.$$ 4. Partial fractions. Write $$\frac{1}{(3u+5)(1+u^2)}=\frac{Au+B}{1+u^2}+\frac{C}{3u+5}.$$ Then $$1=(Au+B)(3u+5)+C(1+u^2).$$ Expanding, $$1=(3A+C)u^2+(5A+3B)u+(5B+C).$$ Comparing coefficients: $$3A+C=0,$$ $$5A+3B=0,$$ $$5B+C=1.$$ From $C=-3A$ and $3B=-5A\Rightarrow B=-\frac{5A}{3}$. Substitute into $5B+C=1$: $$5\left(-\frac{5A}{3}\right)-3A=1 \Rightarrow -\frac{25A}{3}-3A=1 \Rightarrow -\frac{34A}{3}=1 \Rightarrow A=-\frac{3}{34}.$$ Thus $$B=\frac{5}{34},\qquad C=\frac{9}{34}.$$ So $$J=\int_0^1\left(\frac{-\frac{3}{34}u+\frac{5}{34}}{1+u^2}+\frac{9/34}{3u+5}\right)du.$$ That is, $$J=\frac{1}{34}\int_0^1\frac{-3u+5}{1+u^2}\,du+\frac{9}{34}\int_0^1\frac{du}{3u+5}.$$ 5. Integrate termwise. First, $$\int \frac{-3u+5}{1+u^2}du=-\frac{3}{2}\ln(1+u^2)+5\tan^{-1}u.$$ Next, $$\int \frac{du}{3u+5}=\frac13\ln(3u+5).$$ Therefore, $$J=\frac{1}{34}\left[-\frac{3}{2}\ln(1+u^2)+5\tan^{-1}u\right]_0^1+ rac{9}{34}\cdot\frac13\left[\ln(3u+5)\right]_0^1.$$ So $$J=\frac{1}{34}\left(-\frac{3}{2}\ln 2+5\cdot\frac{\pi}{4}\right)+\frac{3}{34}(\ln 8-\ln 5).$$ Since $\ln 8=3\ln 2$, $$J=\frac{5\pi}{136}-\frac{3\ln 2}{68}+\frac{9\ln 2}{34}-\frac{3\ln 5}{34}.$$ Now combine the $\ln 2$ terms: $$-\frac{3}{68}+\frac{18}{68}=\frac{15}{68}.$$ Hence $$J=\frac{5\pi}{136}+\frac{15}{68}\ln 2-\frac{3}{34}\ln 5.$$ 6. Substitute into $I$. $$I=\frac{34\pi}{3}-\frac{680}{3}\left(\frac{5\pi}{136}+\frac{15}{68}\ln 2-\frac{3}{34}\ln 5\right).$$ Now simplify each term: $$\frac{680}{3}\cdot \frac{5\pi}{136}=\frac{25\pi}{3},$$ $$\frac{680}{3}\cdot \frac{15}{68}=50,$$ $$\frac{680}{3}\cdot \frac{3}{34}=20.$$ Thus $$I=\frac{34\pi}{3}-\frac{25\pi}{3}-50\ln 2+20\ln 5.$$ So $$I=3\pi-50\ln 2+20\ln 5.$$ 7. Compare with options. This matches: $$\boxed{\text{A: }3\pi-50\log_e 2+20\log_e 5}.$$

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