Question

Integration of sin x /(4+cos^2 x)(2-sin^2 x)

Answer 1

Answer:

Refer to the attachment

Step-by-step explanation:

Answer 2

Rule:

$\mathsf{\int \frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\:tan^{-1}\frac{x}{a}+C}$

where C is constant of integration

Solution:

Now, $\mathsf{\int \frac{sinx\:dx}{(4+cos^{2}x)(2-sin^{2}x)}}$

$\mathsf{=\int \frac{sinx\:dx}{(4+cos^{2}x)\{2-(1-cos^{2}x)\}}}$

$\mathsf{=\int \frac{sinx\:dx}{(cos^{2}x+4)(cos^{2}x+1)}}$

$\mathsf{=-\int \frac{(-sinx)\:dx}{(cos^{2}x+4)(cos^{2}x+1)}}$

$\mathsf{=-\int \frac{d(cosx)}{(cos^{2}x+4)(cos^{2}x+1)}}$

$\mathsf{=-\frac{1}{3}\:\int \frac{(cos^{2}x+4)-(cos^{2}x+1)}{(cos^{2}x+4)(cos^{2}x+1)}d(cosx)}$

$\mathsf{=-\frac{1}{3}\:\int \frac{d(cosx)}{cos^{2}x+1^{2}}+\frac{1}{3}\:\int \frac{d(cosx)}{cos^{2}x+2^{2}}}$

$\small \mathsf{=-\frac{1}{3}\:tan^{-1}(cosx)+\frac{1}{3}.\frac{1}{2}\:tan^{-1}(\frac{cosx}{2})+C}$

where C is constant of integration

$\small \mathsf{=-\frac{1}{3}\:tan^{-1}(cosx)+\frac{1}{6}\:tan^{-1}(\frac{cosx}{2})+C}$ ,

which is the required integral.