What is #f(x) = int xsinx^2 + tan^2x -cosx dx# if #f(0)=-1 #?

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Answer 1 · 2016-06-17T12:01:01

#f(x)=-1/2cosx^2+tanx-x-sinx+1/2.#

given that, #f(x)=int[xsinx^2+tan^2x-cosx]dx=intxsinx^2dx+inttan^2xdx-intcosdx=I_1+I_2-I_3, say.#

Then, #I_1=intxsinx^2dx, =1/2intsinx^2*2xdx, =1/2intsintdt,# [where #x^2=t# #rArr# #2xdx=dt#]
#=1/2(-cost)=-1/2cosx^2.#

#I_2=inttan^2xdx=int(sec^2x-1)dx=tanx-x.#

#I_3=intcosxdx=sinx.#

#f(x)=I_1+I_2-I_3=-1/2cosx^2+tanx-x-sinx+C.#

But, given that #f(0)=-1.#

Hence, #-1/2cos0+tan0-0-sin0+C=0.# #:. C=1/2.

Finally,

#f(x)=-1/2cosx^2+tanx-x-sinx+1/2.#