Evaluate the following integrals : ∫sin^3x/√cosx dx ​

Question

Evaluate the following integrals :

\(\int\frac{sin^3x}{\sqrt{cosx}}\)dx

Answer 1

In this equation, 

We can manipulate numerator 

sin³x = sin²x.sinx 

∴ Now the equation becomes,

⇒ \(\int\frac{sin^2x.sinx}{\sqrt{cosx}}\)

sin²x = 1 - cos²x

⇒ \(\int\frac{(1-cos^2x).sinx}{\sqrt{cosx}}\)

Now, 

Let us assume cosx = t 

d(cosx) = dt 

- sinx dx = dt 

Substitute values of t and dt in above equation

Comments

In your 7th step you have to put 1-cos^2 x in a bracket.

Answer 2

HELLO,

\(I=\int \frac{\sin^3x}{\sqrt{\cos x}}dx\)

\(I=\int \frac{\sin^2x * \sin x}{\sqrt{\cos x}}dx\)

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

Let \(cosx=t^2\)

\(-\sin x dx =2tdt\)

\(I=\int \frac{1-t^4}{\sqrt{t^2}}2t(-dt)\)

\(I=-\int \frac{1-t^4}{t}2tdt\)

\(I=-2\int(1-t^4)dt\)

\(I=-2(t-t^5/5) + C\)

As \(t=\sqrt{\cos x}\)

\(I= -2(\cos x)^{1/2}+2\frac{(\cos x)^{5/2}}{5} +C\)

I HOPE YOU WILL UNDERSTAND.