What is $f (x) = \int x {sin x}^{2} + {tan}^{2} x - cos x d x$ $f(x) = int xsinx^2 + tan^2x -cosx dx$ if $f (π) = - 2$ $f(pi)=-2$ ?

Question

What is $f (x) = \int x {sin x}^{2} + {tan}^{2} x - cos x d x$ $f(x) = int xsinx^2 + tan^2x -cosx dx$ if $f (π) = - 2$ $f(pi)=-2$ ?

2 Answers

Impact of this question

2120 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-21T14:04:04

$f (x) = tan x - \frac{1}{2} {cos x}^{2} - x - sin x - 2 + \frac{1}{2} {cos π}^{2} + π$ $f(x)=tanx-1/2cosx^2-x-sinx-2+1/2cospi^2+pi$

Explanation:

$f (x) = \int (x sin (x^{2}) + {tan}^{2} x - cos x) d x = \frac{1}{2} \int 2 x sin (x^{2}) d x + \int {sec}^{2} x d x - \int 1 d x - \int cos x d x = - \frac{1}{2} {cos x}^{2} + tan x - x - sin x + c$ $f(x)=int(xsin(x^2)+tan^2x-cosx)dx=1/2int2xsin(x^2)dx+intsec^2xdx-int1dx-intcosxdx=-1/2cosx^2+tanx-x-sinx+"c"$

We are also told

$f (π) = - \frac{1}{2} {cos π}^{2} + tan π - π - sin π + c = - \frac{1}{2} {cos π}^{2} - π + c = - 2 \Rightarrow c = - 2 + \frac{1}{2} {cos π}^{2} + π$ $f(pi)=-1/2cospi^2+tanpi-pi-sinpi+"c"=-1/2cospi^2-pi+"c"=-2 rArr c=-2+1/2cospi^2+pi$

So

$f (x) = tan x - \frac{1}{2} {cos x}^{2} - x - sin x - 2 + \frac{1}{2} {cos π}^{2} + π$ $f(x)=tanx-1/2cosx^2-x-sinx-2+1/2cospi^2+pi$

Answer 2 · 2018-03-21T14:07:40

$f (x) = - \frac{1}{2} {cos x}^{2} + tan x - x - sin x + (π - 2 - \frac{1}{2} {cos π}^{2})$ $f(x)=-1/2cosx^2+tanx-x-sinx+(pi-2-1/2cospi^2)$

Explanation:

Let
$I = \int (x {sin x}^{2} + {tan}^{2} x - cos x) d x$ $I=int(xsinx^2+tan^2x-cosx)dx$
By sum rule

$I = \int (x {sin x}^{2}) d x + \int ({tan}^{2} x) d x + \int (cos x) d x$ $I=int(xsinx^2)dx+int(tan^2x)dx+int(cosx)dx$
Let
$I_{1} = \int (x {sin x}^{2}) d x$ $I_1=int(xsinx^2)dx$

$I_{2} = \int ({tan}^{2} x) d x$ $I_2=int(tan^2x)dx$

$I_{3} = \int (cos x) d x$ $I_3=int(cosx)dx$

$I = I_{1} + I_{2} - I_{3}$ $I=I_1+I_2-I_3$

Now,

$I_{1} = \int (x {sin x}^{2}) d x$ $I_1=int(xsinx^2)dx$

Rearranging

$I_{1} = \int ({sin x}^{2}) (x d x)$ $I_1=int(sinx^2)(xdx)$

Let
$u = x^{2}$ $u=x^2$

${sin x}^{2} = sin u$ $sinx^2=sinu$

$\frac{d u}{d x} = 2 x$ $(du)/dx=2x$

$x d x = \frac{1}{2} d u$ $xdx=1/2du$

$I_{1} = \int sin u (\frac{1}{2} d u)$ $I_1=intsinu(1/2du)$

$I_{1} = - \frac{1}{2} \int (- sin u d u)$ $I_1=-1/2int(-sinudu)$

$\int - sin u d u = cos u$ $int-sinudu=cosu$

$u = x^{2}$ $u=x^2$

$I_{1} = - \frac{1}{2} {cos x}^{2}$ $I_1=-1/2cosx^2$

$I_{2} = \int ({tan}^{2} x) d x$ $I_2=int(tan^2x)dx$

${sec}^{2} x - {tan}^{2} x = 1$ $sec^2x-tan^2x=1$

${tan}^{2} x = {sec}^{2} x - 1$ $tan^2x=sec^2x-1$

$\int ({tan}^{2} x) d x = \int ({sec}^{2} x - 1) d x$ $int(tan^2x)dx=int(sec^2x-1)dx$

$= \int {sec}^{2} x d x - \int 1 d x$ $=intsec^2xdx-int1dx$

$\int {sec}^{2} x d x = tan x$ $intsec^2xdx=tanx$

$\int 1 d x = x$ $int1dx=x$

$\int ({tan}^{2} x) d x = tan x - x$ $int(tan^2x)dx=tanx-x$

$I_{2} = tan x - x$ $I_2=tanx-x$

$I_{3} = \int (cos x) d x$ $I_3=int(cosx)dx$

$\int cos x d x = sin x$ $intcosxdx=sinx$

$I_{3} = sin x$ $I_3=sinx$

$I = I_{1} + I_{2} - I_{3}$ $I=I_1+I_2-I_3$

$I = \int (x {sin x}^{2} + {tan}^{2} x - cos x) d x$ $I=int(xsinx^2+tan^2x-cosx)dx$

$I_{1} = - \frac{1}{2} {cos x}^{2}$ $I_1=-1/2cosx^2$

$I_{2} = tan x - x$ $I_2=tanx-x$

$I_{3} = sin x$ $I_3=sinx$

$\int (x {sin x}^{2} + {tan}^{2} x - cos x) d x = - \frac{1}{2} {cos x}^{2} + tan x - x - sin x$ $int(xsinx^2+tan^2x-cosx)dx=-1/2cosx^2+tanx-x-sinx$

Also

$f (π) = - 2$ $f(pi)=-2$

Here,

$f (x) = \int (x {sin x}^{2} + {tan}^{2} x - cos x) d x + C$ $f(x)=int(xsinx^2+tan^2x-cosx)dx+C$

$f (π) = - \frac{1}{2} {cos π}^{2} + tan π - π - sin π + C$ $f(pi)=-1/2cospi^2+tanpi-pi-sinpi+C$

$tan π = 0$ $tanpi=0$

#sinpi=0

$- 2 = - \frac{1}{2} {cos π}^{2} - π + C$ $-2=-1/2cospi^2-pi+C$ #

$C = π - 2 - \frac{1}{2} {cos π}^{2}$ $C=pi-2-1/2cospi^2$

$f (x) = - \frac{1}{2} {cos x}^{2} + tan x - x - sin x + (π - 2 - \frac{1}{2} {cos π}^{2})$ $f(x)=-1/2cosx^2+tanx-x-sinx+(pi-2-1/2cospi^2)$

Science

Math

Humanities

... and beyond

What is $f (x) = \int x {sin x}^{2} + {tan}^{2} x - cos x d x$ $f(x) = int xsinx^2 + tan^2x -cosx dx$ if $f (π) = - 2$ $f(pi)=-2$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is f(x) = int xsinx^2 + tan^2x -cosx dxf(x)=∫xsinx2+tan2x−cosxdxf(x) = int xsinx^2 + tan^2x -cosx dx if f(pi)=-2 f(π)=−2f(pi)=-2 ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

What is $f (x) = \int x {sin x}^{2} + {tan}^{2} x - cos x d x$ $f(x) = int xsinx^2 + tan^2x -cosx dx$ if $f (π) = - 2$ $f(pi)=-2$ ?