# Get Answers to all your Questions

### Answers (1)

Answer:$\sin ^{-1}\left[\frac{1}{3}(\sin x+\cos x)\right]+c$

Hint  $\int \frac{d t}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}$

Given: $\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x$

Explanation:

$\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x$

$=\int \frac{\cos x-\sin x}{\sqrt{8+1-1-\sin 2 x}} d x$

$=\int \frac{\cos x-\sin x}{\sqrt{9-\left(\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\right)}} d x \quad\left[\begin{array}{l} \sin ^{2} x+\cos ^{2} x=1 \\ \sin 2 x=2 \sin x \cdot \cos x \end{array}\right]$

$=\int \frac{\cos x-\sin x}{\sqrt{9-(\sin x+\cos x)^{2}}} d x$

Let $\cos x+\sin x=t$

$(-\sin x+\cos x) d x=d t$

Put in (1)

$\int \frac{d t}{\sqrt{3^{2}-t^{2}}}=\sin ^{-1}\left[\frac{t}{3}\right]+c$                                                    $\left[\int \frac{d t}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}\right]$

$=\sin ^{-1}\left[\frac{1}{3}(\sin x+\cos x)\right]+c$

View full answer

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support