IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

nikunj_vm

Solve

nivedh_89

ans1....)
let, x/2+

/4 = m.............

let the required integral = I = -e^x*cosm+e^x*sinm - _[0]

^[2] e^x*sinm

the last term is nothing but I............
therefore,2I = -e^x*cosm+e^x*sinm
= 0............bcas when we substitute either 0 or 2

the total comes out to be 0

so,I = 0/2 = 0.................

hence .........the answer of 1st question = 0.............

neeraj_agarwal_1990

Thnx for posting the questions
mujhe bhi inmein problem thi...

jatinroxx

Nivedh, there's a serious mistake.

When u substitute, the limit changes.

jatinroxx

Try making a graph of this fn. You'll realize ur mistake.

The answer can never be 0

Solitude

the ans. to the 1st Q is

{4sin(pi/4+x/2).e^x-2cos(pi/4+x/2)}/5

solve by integration by parts.

jatinroxx

Apply by parts twice..

The integral on the RHS resembles the initial function...

The problem is solved...

jatinroxx

Is the answer

2/5 * (e^2pi + 1)

i.e. within the limits of human error...

jatinroxx

For the second question,

First substitute

ln(x) = t^2

This gets away with the radical sign as well as the limits become 1 -> 2

Next is regular solving.

Just have to manipulate the integral to bring it to the given form...

nivedh_89

but what was the necessity of boink-ing me..........oh! cmmon yaar!!!!!!!!!!!