Question

solve it in simple way..
integration!!

Answer 1

Hi,

Here is your answer !
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Let , ɪ = ʃ tan⁴x dx

= ʃ tan²x . tan²x dx

= ʃ ( sec²x − 1 ) . tan²x dx

= ʃ ( sec²x . tan²x − tan²x ) dx

= ʃ ( sec²x . tan²x ) dx − ʃ tan²x dx

= ɪ₁ − ɪ₂ ( say )

Now ,

ɪ₁ = ʃ ( sec²x . tan²x ) dx

Let t = tanx then dt = sec²x dx

→ ɪ₁ = ʃ t² dt

→ ɪ₁ = t³/3 + C₁

→ ɪ₁ = tan³x/3 + C₁

And ,

ɪ₂ = ʃ tan²x dx = ʃ (sec²x − 1) dx

ɪ₂ = ʃ sec²x dx − ʃ 1 dx

ɪ₂ = tanx − x + C₂

Now,

ɪ = ɪ₁ − ɪ₂

ɪ = tan³x/3 + C₁ − tanx + x − C₂

ɪ = tan³x/3 − tanx + x + C

( Where, C = C₁ − C₂ )

Hence,

ʃ tan⁴x dx = tan³x / 3 − tanx + x + C

Answer 2

The answer is explained below..


Hope it helps!