Question

(3 senx + 4 cosec x)? d​

Answer 1

Step-by-step explanation:

We have,

$\int(3 \sin(x) + 4 \cosec(x))^{2} dx$

$= \int \{9 \sin^{2} (x) + 16 \cosec^{2} (x) + 24 \} dx$

$= \int 9 \sin^{2} (x)dx + \int 16 \cosec^{2} (x) dx+ \int24 dx$

$= 9\int \sin^{2} (x)dx +16 \int \cosec^{2} (x) dx+ 24 \int dx$

$= 9\int \frac{1 - \cos(2x) }{2} dx - 16 \cot (x) + 24 x +C$

$= \frac{9}{2} \int \{1 - \cos(2x) \} dx - 16 \cot (x) + 24 x +C$

$= \frac{9}{2} \int dx - \frac{9}{2} \int\cos(2x) dx - 16 \cot (x) + 24 x +C$

$= \frac{9x}{2} - \frac{9}{4} \sin(2x) - 16 \cot (x) + 24 x +C$

$= \frac{9x}{2} + 24x - \frac{9}{4} \sin(2x) - 16 \cot (x) +C$

$= \frac{57x}{2} - \frac{9}{4} \sin(2x) - 16 \cot (x) +C$