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)#