Question

Write a value of ∫1−tanxx+logcosxdx

Answer 1

Answer:

(1/2) ( x + logcosx)² + c

Step-by-step explanation:

To find --->

∫ ( 1 - tanx ) ( x + logCosx ) dx

Solution--->

I= ∫ ( 1 - tanx ) (x + logCosx ) dx

Let x + logCosx = t

We have some formulee

d/dx ( x ) = 1

d/dx ( logx )= 1/x

d/dx ( Cosx ) = -Sinx

Applying it here

Differentiating with respect to x we get

{1 + ( 1 / Cosx ) ( - Sinx )} dx = dt

{ 1 - (Sinx / Cosx) } dx = dt

( 1 - tanx ) dx = dt

Now

I = ∫ t dt

We have a formula

∫ xⁿ dx = xⁿ⁺¹ / ( n+1 ) + c

Applying it here we get

= t² / 2 + c

= (1/2) ( x + logCosx )² + c