Question

The value of $\int \frac{co{s}^{3}x}{si{n}^{2}x+sinx}dx$ is equal to

• A. $lo{g}_e\left|sinx\right|+sinx+C$
• B. $lo{g}_e\left|sinx\right|-sinx+C$
• C. $-lo{g}_e\left|sinx\right|=-sinx+C$
• D. $-lo{g}_e\left|sinx\right|+sinx+C$

Answer 1

The correct option is B $lo{g}_e\left|sinx\right|-sinx+C$

Let $I=\int \frac{{\cos }^{3}x}{{\sin }^{2}x+\sin x}dx$

$I=\int \frac{{\cos }^{2}x.cosx}{{\sin }^{2}x+\sin x}dx$

$I=\int \frac{(1-{\sin }^{2}x).cosx}{{\sin }^{2}x+\sin x}dx$

Let $\sin x=t$

$\cos xdx=dt$

$I=\int \frac{(1-t^2)}{t^2+t}dt$

$I=\int \frac{(1-t)(1+t)}{t(1+t)}dt$

$I=\int \frac{(1-t)}{t}dt$

$I=\int \frac{1}{t}dt-\int \left(1\right)dt$

$I={\log }_e|t|-t+c$

Now, re-substituting value of $t,$ we get:

$I={\log }_e|\sin x|-\sin x+c$