Q. 12

# Evaluate

We can write ∫cos33xdx as:

∫cos3x(cos23x)dx and

further as:

=cos3x(1-sin23x)dx

=∫cos3xdx-∫cos3x(sin23x)dx

Taking A=∫cos3xdx

Solving for A

A=

Taking B=∫cos3x(sin23x)dx

In this taking sin3x=t

Differentiating on both sides we get

3cos3xdx=dt

Solving by putting these values we get

B

Substituting values we get

B

Our final answer is A+B i.e

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