Question

Integrate cos2x/(cosx+sinx)^2

Answer 1

$\textbf{Given:}$

$\displaystyle\int{\frac{cos2x}{(cosx+sinx)^2}}\;dx$

$=\displaystyle\int{\frac{cos2x}{cos^2x+sin^2x+2\,sinx\,cosx}}\;dx$

$=\displaystyle\int{\frac{cos2x}{1+sin2x}}\;dx$

$\text{We apply change of variable method to solve the integral}$

$\boxed{\begin{minipage}{3cm}\textbf{Take t=1+sin2x}\\\\ \frac{dt}{dx}=2\;cos2x\\\\\implies\,\frac{dt}{2}=cos2x\,dx \end{minipage}}$

$=\displaystyle\int{\frac{1}{t}}\;\frac{dt}{2}$

$=\displaystyle\frac{1}{2}\int{\frac{1}{t}}\;dt$

$=\displaystyle\frac{1}{2}\,log\,t+c$

$=\displaystyle\frac{1}{2}\,log(1+sin2x)+c$

$\therefore\bf\displaystyle\int{\frac{cos2x}{(cosx+sinx)^2}}\;dx=\frac{1}{2}\,log(1+sin2x)+c$

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Answer 2

Answer:

there are two option

Step-by-step explanation:

hope it help you