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

2 Answers
Mar 21, 2018

f(x)=tanx-1/2cosx^2-x-sinx-2+1/2cospi^2+pi

Explanation:

f(x)=int(xsin(x^2)+tan^2x-cosx)dx=1/2int2xsin(x^2)dx+intsec^2xdx-int1dx-intcosxdx=-1/2cosx^2+tanx-x-sinx+"c"

We are also told

f(pi)=-1/2cospi^2+tanpi-pi-sinpi+"c"=-1/2cospi^2-pi+"c"=-2 rArr c=-2+1/2cospi^2+pi

So

f(x)=tanx-1/2cosx^2-x-sinx-2+1/2cospi^2+pi

Mar 21, 2018

f(x)=-1/2cosx^2+tanx-x-sinx+(pi-2-1/2cospi^2)

Explanation:

Let
I=int(xsinx^2+tan^2x-cosx)dx
By sum rule

I=int(xsinx^2)dx+int(tan^2x)dx+int(cosx)dx
Let
I_1=int(xsinx^2)dx

I_2=int(tan^2x)dx

I_3=int(cosx)dx

I=I_1+I_2-I_3

Now,

I_1=int(xsinx^2)dx

Rearranging

I_1=int(sinx^2)(xdx)

Let
u=x^2

sinx^2=sinu

(du)/dx=2x

xdx=1/2du

I_1=intsinu(1/2du)

I_1=-1/2int(-sinudu)

int-sinudu=cosu

u=x^2

I_1=-1/2cosx^2

I_2=int(tan^2x)dx

sec^2x-tan^2x=1

tan^2x=sec^2x-1

int(tan^2x)dx=int(sec^2x-1)dx

=intsec^2xdx-int1dx

intsec^2xdx=tanx

int1dx=x

int(tan^2x)dx=tanx-x

I_2=tanx-x

I_3=int(cosx)dx

intcosxdx=sinx

I_3=sinx

I=I_1+I_2-I_3

I=int(xsinx^2+tan^2x-cosx)dx

I_1=-1/2cosx^2

I_2=tanx-x

I_3=sinx

int(xsinx^2+tan^2x-cosx)dx=-1/2cosx^2+tanx-x-sinx

Also

f(pi)=-2

Here,

f(x)=int(xsinx^2+tan^2x-cosx)dx+C

f(pi)=-1/2cospi^2+tanpi-pi-sinpi+C

tanpi=0

#sinpi=0

-2=-1/2cospi^2-pi+C#

C=pi-2-1/2cospi^2

f(x)=-1/2cosx^2+tanx-x-sinx+(pi-2-1/2cospi^2)