What is f(x) = int xsinx^2 + tan^2x -cosx dxf(x)=∫xsinx2+tan2x−cosxdx if f(0)=-1 f(0)=−1?

1 Answer
Ratnaker Mehta
Jun 17, 2016

f(x)=-1/2cosx^2+tanx-x-sinx+1/2.f(x)=−12cosx2+tanx−x−sinx+12.

Explanation:

given that, f(x)=int[xsinx^2+tan^2x-cosx]dx=intxsinx^2dx+inttan^2xdx-intcosdx=I_1+I_2-I_3, say.f(x)=∫[xsinx2+tan2x−cosx]dx=∫xsinx2dx+∫tan2xdx−∫cosdx=I1+I2−I3,say.

Then, I_1=intxsinx^2dx, =1/2intsinx^2*2xdx, =1/2intsintdt,I1=∫xsinx2dx,=12∫sinx2⋅2xdx,=12∫sintdt, [where x^2=tx2=t rArr⇒ 2xdx=dt2xdx=dt]
=1/2(-cost)=-1/2cosx^2.=12(−cost)=−12cosx2.

I_2=inttan^2xdx=int(sec^2x-1)dx=tanx-x.I2=∫tan2xdx=∫(sec2x−1)dx=tanx−x.

I_3=intcosxdx=sinx.I3=∫cosxdx=sinx.

f(x)=I_1+I_2-I_3=-1/2cosx^2+tanx-x-sinx+C.f(x)=I1+I2−I3=−12cosx2+tanx−x−sinx+C.

But, given that f(0)=-1.f(0)=−1.

Hence, -1/2cos0+tan0-0-sin0+C=0.−12cos0+tan0−0−sin0+C=0. #:. C=1/2.

Finally,

f(x)=-1/2cosx^2+tanx-x-sinx+1/2.f(x)=−12cosx2+tanx−x−sinx+12.

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