Question

Integrate x ( sinx) / ( 1 + cos2x) ?

Answer 1

hope this help you friend

Answer 2

Answer:

$\frac{1}{2} [xSecx- ln(Secx-Tanx)]+c$

Step-by-step explanation:

We have to evaluate, ∫$\frac{xSinx}{1+Cos2x}$dx.

Now,  ∫$\frac{xSinx}{1+Cos2x}$dx.

= ∫$\frac{xSinx}{2Cos^{2}x } dx$ {Since, we have the formula Cos2x=2Cos²x -1 }

= $\frac{1}{2}$∫xTanx Sec x dx

Here we will apply the formula, ∫u.vdx=u∫v.dx-∫[(du/dx)(∫v.dx)]dx and here u= x and v= tanx Secx.

=$\frac{1}{2}[xSecx-\int\ {Secx} \, dx ]$

= $\frac{1}{2} [xSecx- ln(Secx-Tanx)]+c$ {Where c is an integration constant} (Answer)