The integral int_0^{frac{pi}{4}} frac{136 sin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} dx$. We express the numerator as $136 \sin x = A(3 \sin x + 5 \cos x) + B(3 \cos x

Vedclass · Accepted Answer

$3 \pi-50 \log _e 2+20 \log _e 5$

Vedclass · Answer

(A) Let $I = \int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} dx$. We express the numerator as $136 \sin x = A(3 \sin x + 5 \cos x) + B(3 \cos x - 5 \sin x)$. Comparing coefficients of $\sin x$ and $\cos x$: $136 = 3A - 5B$ ... $(1)$ $0 = 5A + 3B$ ... $(2)$ From $(2)$,$B = -\frac{5}{3}A$. Substituting into $(1)$: $136 = 3A - 5(-\frac{5}{3}A) = 3A + \frac{25}{3}A = \frac{34}{3}A$. Thus,$A = \frac{136 	imes 3}{34} = 12$ and $B = -\frac{5}{3}(12) = -20$. Now,$I = \int_0^{\pi / 4} \frac{12(3 \sin x + 5 \cos x) - 20(3 \cos x - 5 \sin x)}{3 \sin x + 5 \cos x} dx$. $I = 12 \int_0^{\pi / 4} dx - 20 \int_0^{\pi / 4} \frac{3 \cos x - 5 \sin x}{3 \sin x + 5 \cos x} dx$. $I = 12[x]_0^{\pi / 4} - 20[\ln|3 \sin x + 5 \cos x|]_0^{\pi / 4}$. $I = 12(\frac{\pi}{4}) - 20[\ln(\frac{3}{\sqrt{2}} + \frac{5}{\sqrt{2}}) - \ln(5)]$. $I = 3\pi - 20[\ln(\frac{8}{\sqrt{2}}) - \ln(5)] = 3\pi - 20[\ln(4\sqrt{2}) - \ln(5)]$. $I = 3\pi - 20[\ln(2^{5/2}) - \ln(5)] = 3\pi - 20[\frac{5}{2}\ln 2 - \ln 5]$. $I = 3\pi - 50 \ln 2 + 20 \ln 5$.

The integral $\int_0^{\frac{\pi}{4}} \frac{136 \sin x}{3 \sin x+5 \cos x} dx$ is equal to :

Similar Questions

$\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}} {\left( {\frac{1}{2}{{(3\sin \theta )}^2} - \frac{1}{2}{{(1 + \sin \theta )}^2}} \right)\,d\theta } $

$\int_0^3 \frac{3x+1}{x^2+9} dx$ is equal to :

$\int_{a-c}^{b-c} f(x+c) \, dx = $

$\int_0^{\pi /3} \cos 3x \, dx = $

$\int_{1}^{5} (|x-3| + |1-x|) dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module