Indefinite Integration 1 Question 85
86. Evaluate $\int _0^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$.
$(1983,3 M)$
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Answer:
Correct Answer: 86. $\frac{1}{20}(\log 3)$
Solution:
- Let $I=\int _0^{\pi / 4} \frac{(\sin x+\cos x)}{9+16 \sin 2 x} d x$
$$ I=\int _0^{\pi / 4} \frac{\sin x+\cos x}{25-16(\sin x-\cos x)^{2}} d x $$
Put $4(\sin x-\cos x)=t \Rightarrow 4(\cos x+\sin x) d x=d t$
$\therefore \quad I=\frac{1}{4} \int _{-4}^{0} \frac{d t}{25-t^{2}}=\frac{1}{4} \cdot \frac{1}{2(5)} \log \frac{5+t}{5-t}$
$$ \begin{aligned} I & =\frac{1}{40} \log \frac{5+0}{5-0}-\log \frac{5-4}{5+4} \\ & =\frac{1}{40} \log 1-\log \frac{1}{9}=\frac{1}{40} \log 9=\frac{1}{20}(\log 3) \end{aligned} $$
