IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

shinee

1)

(tan⁹ x)dx/(sinxcosx)
2)

sin^5/2xcos³xdx
3)

(1-sin2x)dx
4)

1dx/(1+e^x )(1-e^-x )
5)

sin²xcos²xdx
6)_{[ ]}

^{[ ]} 1/(x⁴ +1)

chimanshu_007

ans 3) just take 1 = sin^2x + cos^2x

and open sin2x = 2sinxcosx

it will become a whole sq den it will be easy

ans 1)

tan⁹x / sinx cosx

= tan⁸x . (sinx)/ sinx . cos²x

= tan⁸x.sec²x

now put tan x = t

ashish_banga

6 )
divide numerator and denominator by x^2 and make algebric twins

chimanshu_007

ans 4) integral dx / (1 + e^x)(1 - e^-x)

= dx / (1 + e^x)(1 - (1/e^x))

= e^xdx / (1 + e^x)(e^x - 1)..................just taking the LCM

= e^xdx / (e^2x - 1)...................(a+b)(a-b) = a² - b²

now put e^x = t , e^xdx = dt

it becomes

dt / t² - 1

now its easy

ashish_banga

5
it is = [ sin(2x) ]^2 / 4 = [ 1 - cos(8x) ] / 8
it is easy afterwards

studyid

4) in the deno. u get e^-x as 1/e^x .....

then take the lcm ........ so u will get overall :

integration of (e^x dx)/(e^(2x) -1)

now put e^x as t ......... and so .....u will get ......

integration of dt/(t^2-1)

now the integral can b calculated easily ,

edit : sorry .....waz busy writing the solution ..... when himanshu bhai already solved it ..

uday_zingtudor

1) You can write it as tan⁸xsec²x

So the ans is tan⁹x/9.

little_genius

1/2{x^2+1/x^4+1 - (x^2-1)/x^4+1}

divide by x^2 both num &denomitor and then express num as derivative of denominator

RyuAmakusa

well i guess only 2 is remaining.
sin5/2xcos3xdx =

sin²x.cos³xsin^1/2x dx. now putsin^1/2x = t

u get 2

t⁶(1-t⁴).dt.now it is easy.

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