Definite Integration Question 84

Question 84

  1. 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:

  1. 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 $4(\sin x-\cos x)=t \Rightarrow 4(\cos x+\sin x) d x=d t$

$\therefore \quad I=\frac{1}{4} \int_{-4}^{0} \frac{d t}{25-t^{2}}=\frac{1}{4} \cdot \frac{1}{2(5)} \log \frac{5+t}{5-t}$

$$ \begin{aligned} I & =\frac{1}{40} \log \frac{5+0}{5-0}-\log \frac{5-4}{5+4} \ & =\frac{1}{40} \log 1-\log \frac{1}{9}=\frac{1}{40} \log 9=\frac{1}{20}(\log 3) \end{aligned} $$



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