Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

The integral ∫xxsinx+cosx2dx is equal to (where C is a constant of integration) :

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

tanx−xsecxxsinx+cosx+C

b

secx+xtanxxsinx+cosx+C

c

secx−xtanxxsinx+cosx+C

d

tanx+xsecxxsinx+cosx+C

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

∫x2(xsinx+cosx)2dx  Multiply and divide by cosx  =∫xcosx·xcosx(xsinx+cosx)2dx Integration by parts=xcosx∫xcosx(xsinx+cosx)2dx-∫ddxxcosx-∫xcosx(xsinx+cosx)2dxdx                                 put xsinx+cosx=t                                ⇒(xcosx+sinx-sinx)dx=dt                                  ⇒ xcosx dx=dt=xcosx·-1(xsinx+cosx)+∫cosx+xsinxcos2x·1(xsinx+cosx)dx=-xcosx(xsinx+cosx)+tanx+C=tanx-xsecxxsinx+cosx+C
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring