Question

substitution method 1/√1-tan²x. dx​

Answer 1

Answer:

Hope this helps you !!!!

Answer 2

$\;\mapsto\;\;\underline{\underline{\bf{\blue{Solution\::-}}}}$

Rewrite the integral as

S cos x cos2 x dx=[cosx1-sin2xdx]

Make the substitution u-sinx. Then du-cosxdx. So now our indefinite integral is

Jdu1-u2.

It is more convenient to make the substitution in the "limits" of integration. When x=π/4, we have u-1/2-√ and when x=0, we have u=0, so we want

Sπ/4u=0du1-u2.

Our integral is now in perfect shape for the method of "partial fractions." By going through the machinery of partial fractions, or by inspection, we have

11-u2-11-u2=1/21-u+1/21+u.

We have arrived at

(1/2 √0(1/21-u+1/21+u)du.

Finally, everything is easy. By substitution, or by inspection, we have

[1/21-udu=-(1/2)In(|1-ul)