IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

Taara

1)1 / ( sqrt ( sin^3 x sin(x+a))

2) integrate 0 to pi/4 sin x + cos x / (9 + 16 sin 2x)

3) integrate 0 to pi xtanx / (sec x + tan x)

4) integrate 0 to pi/2 log ( (4 + 3 sin x) / (4 + 3 cos x))

5) -5 to 5 | x + 2 |

6) e^(2x) sin x

sboosy

(-5 to 5)

|x+2| dx

x+2 >= 0 for all x>=-2

so splitting into two we get

(-5 to -2)

-(x+2) dx + (-2 to 5)

(x+2) dx

so we get

[(x+2)²/2 ] (-2 to -5) + [(x+2)²/2 ] (-2 to 5)

so

9/2 + 49/2 = 58/2 = 29

anchitsaini

for
3)multiply numerator and denominator by (sec x - tan x)
then denominator would become 1 as (sec^2 x - tan^2 x) =1

and it easy to solve the integral now

sboosy

3) integral xtanx/ (secx+tanx) dx
so xtanx(secx-tanx)/(sec^2x -tan^2 x)
now (sec^2x -tan^2 x) = 1
xtanxsecx dx - xtan^2x
x(d secx) - x (sec^2x-1) dx
x secx - log|secx+tanx| - xtanx + log|secx| +x^2/2 + c

rishab_09

For Qs no 6:

Integrate it with the help of integration by parts rule

Take e^2x as 1st function and sinx as 2nd function( ILATE rule)

I= e^2x

sinxdx-

{d/dx(e^2x)

sinxdx}dx

=e^2x.(-cosx)+

2e^2x.cosxdx

=-cosx.e^2x+2

e^2x.cosxdx

=-cosx.e^2x+2I(1)

Now again using parts rule for I(1)

I(1)=e^2xcosxdx-{d/dx(e^2x)cosxdx}dx

=sinx.e^2x-2e^2x.sinxdx

=sinx.e^2x-2I

Therefore,I= -cosx.e^2x+2sinx.e^2x-4I

or, 5I=-cosx.e^2x+2sinx.e^2x

or, I=(-cosx.e^2x+2sinx.e^2x)/5+c

Hope u find it useful.....

raulrag009

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