CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

Find the integral of the function
cosxsinx1+sin2x

Open in App
Solution

cosxsinx1+sin2x

=cosxsinxsin2x+cos2x+2sinxcosx

=cosxsinx(cosx+sinx)2

On integrating, we get

cosxsinx(cosx+sinx)2dx

Let cosx+sinx=t --(1)

Then,

ddx(cosx+sinx)=ddxt

sinx+cosx=dtdx

(cosxsinx)dx=dt---(2)

Now,

cosxsinx(cosx+sinx)2dx

=dtt2

=t2dt [Since, from (1) and (2)]

=t2+12+1+C [Since, xn=xn+1n+1,n1]

=1t+C

=1cosx+sinx+C [Where, C is any real constant]

cosxsinx(cosx+sinx)2dx=1cosx+sinx+C


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Integration of Trigonometric Functions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon