Question

integrate (x.sinx) / ( (cos x) cube)

Answer 1

Answer:

(x/2) Sec²x - (1/2) tanx + c

Step-by-step explanation:

Let

I = ∫ x Sinx / Cos³x dx

= ∫ x (Sinx / Cosx) (1/ Cos²x) dx

= ∫ x tan x sec² x dx

Let tan x = t => x = tan⁻¹ t

differentiating both sides

sec² x dx = dt

I = ∫ tan⁻¹ t t dt

Now applying intregation by parts

= tan⁻¹ t (t²/ 2) - ∫ (1 / 1+t²) (t² / 2) dt

= (t² tan⁻¹t )/ 2 -1/2 ∫ (t² /1+t² ) dt

( 1+t² )- 1

=(t² tan⁻¹t)/2 -1/2 ∫ ----------------- dt

( 1+t²)

t² tan⁻¹ t 1+t² 1

= -------------- -1/2 ∫ (---------- - ---------) dt

2 1+t² 1+t²

t² tan⁻¹ t 1

= --------------- - 1/2 ∫ ( 1 - ------------) dt

2 1+ t²

t² tan⁻¹ t

= ---------------- - 1/2 ( t - tan⁻¹ t ) + c

2

t² tan⁻¹t t tan⁻¹t

= -------------- - ------- + ------------- + c

2 2 2

= 1/2 (t² + 1) tan⁻¹t - 1/2 t + c

= 1/2 ( tan²x + 1) x - 1/2 tanx + c

= (x/2 ) Sec²x -( 1/2 ) tanx + c