Question

Evaluate ∫ sinx/cos2x dx

\(\int\frac{sin\,x}{cos\,2x}\) dx

Answer 1

Let I = ∫ sinx/cos2x dx

We know, 

cos 2x = 2cos²x – 1 

Putting values in I we get,

I = \(\int\frac{sin\,x}{cos\,2x}\) dx

= \(\int\frac{sin\,x}{2\,cos^2x-1}\)dx

Put cosx = t 

Differentiating w.r.t to x we get, 

sinx dx = - dt 

Putting values in integral we get,

I = \(-\int\frac{dt}{2t^2-1}\)

= \(-\int\frac{dt}{(\sqrt 2t)^2-(1)^2}\)

Again,

Put √2× t = u 

Differentiating w.r.t to t we get,

dt = \(\frac{du}{\sqrt 2}\)

Putting values in integral we get,

I = \(\frac{1}{\sqrt 2}\)\(\int\frac{du}{(1)^2-(u)^2}\)

We know,

\(\int\frac{du}{(1)^2-(x)^2}\) = sin^-1x + c

Substituting value of u we get,

I = \(\frac{1}{\sqrt 2}\)sin^-1\(\sqrt 2\) t + c

Substituting value of t we get,