Question

Integration of (Sinx+cosx)/(9+16sin2x)dx

Answer 1

Please check the attachment. Thank you.

Answer 2

$\bold{\left(\frac{\sin x+\cos x}{9+16 \sin 2 x}\right) d x=\frac{1}{40} \log \left[\frac{(5+4 \sin x-4 \cos x)}{5-4 \sin x+4 \cos x}\right]}$

Step-by-step explanation:

Let,

$I=\frac{\sin x+\cos x}{9+16 \sin 2 x}$

We know that,

$\sin ^{2} x+\cos ^{2} x=1 ; \sin ^{2} x=2 \sin x \cdot \cos x$

$(\sin x-\cos x)^{2}=\sin ^{2} x+\cos ^{2} x-2 \sin x \cdot \cos x=1-\sin 2 x \rightarrow(1)$

$I=\frac{\sin x+\cos x}{9+16-16+16 \sin 2 x}$

$I=\frac{\sin x+\cos x}{25-16(1-\sin 2 x)}$

From (1)

$I=\frac{\sin x+\cos x}{25-16(\sin x-\cos x)^{2}}$

Consider integral on both sides with x

$\int I d x=\int \frac{(\sin x+\cos x)}{25-16(\sin x-\cos x)^{2}} d x$

Let

$sinx - cosx = t$

Differentiating both sides with x

$\frac{d}{d x}(\sin x-\cos x)=\frac{d t}{d x}$

$\cos x-(-\sin x)=\frac{d t}{d x}$

$\cos x+\sin x=\frac{d t}{d x}$

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

$\int I d x=\int \frac{d t}{25-16 t^{2}}$

$\int I d x=\int \frac{d t}{\frac{25}{16}-t^{2}}$

$\int I d x=\frac{1}{16}\left[\frac{1}{2\left(\frac{5}{4}\right)} \log \left(\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right)\right]$

$\int I d x=\frac{1}{16}\left[\left(\frac{4}{10}\right) \log \left(\frac{5+4 t}{5-4 t}\right)\right]$

$\int I d x=\frac{1}{40} \log \left[\frac{(5+4 \sin x-4 \cos x)}{5-4 \sin x+4 \cos x}\right]$