Integration

∫{sin2x}/{(cos4x)+(sin4x)}dx

Pls post full solution [1]

6 Answers

1
krish1092 ·

Write sin2x as 2sinxcosx,take cos^4x common from the denominator,We'll get the integrand as ∫2tanxsec2x/(1+tan4x) now put tanx=t
==>∫ 2tdt/(1+t4) ,put t2=u,Now it will be simple to solve

1
skygirl ·

brilliant krish :)

this is in ncert naa ?

1
skygirl ·

that ncert miscellaneous exercise is really worth doing.... atleast i felt so...

1
greatvishal swami ·

yes skygirl & it was the only chapter which i felt like doing frm ncert

11
Anirudh Narayanan ·

Thanx Krish [1]

341
Hari Shankar ·

or cos4x + sin4x = (cos2x+sin2x)2 - 2 sin2x cos2x

= 1 - sin22x/2 = (1+cos22x)/2

Now put cos 2x = t and integrate

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