IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

shinee

can u plz integrate these
1)sec⁵xtanx    ans    1/5(sec⁵x)
2)(cos2x)/(sinx +cosx)² =   ans   log lsinx+cosxl
3)(x^e-1+ e^x-1 )/(x^e+   e^x ) ans 1/e{logIx^e +e^xl}
can u also tell me how(on wat basis) we should take substitutions like t,z etc
4)(sin2x)/(sin5xsin3x) ans 1/3LogIsin3xI-1/5logIsinxI
5)(1-tanx)/(1+tanx)     ans log Icosx+sinxI
6)cot³x ans -1/2cot²x - log lsinxl
7)(sin2x)/(sin5xsin3x)   ans   1/3loglsin3xl - 1/5 loglsin5xl
8)1/(a²sin²x + b²cos²x) ans 1/ab(tan^-1(atanx/b)
9)1/(e^x + e^-x )²   ans -1/(2(1+e^2x ))
10) 1/(e^x +1)(1+e^-x )   ans -1/(1+e^x )
11)(x+2)/

(1-x² ) ans -

(1-x² ) + 2sin^-x
12)cosec⁴x ans -cotx -(cot³x)/3
13) tan³x ans -loglsecxl +(tan² x)/2

varshavallig

1)i=sec^4x secxtanx put t=secx dt =secxtanx

2)cos2x=cos^2x - sin^2x
substitute and put t=sinx +cosx dt=cosx-sinx

3)put t=x^e+e^x dt=e(x^e-1+e^x-1)

5)tanx=sinx/cosx substitute,proceed same as 2nd
!!!!!!!!!!!!!!!

varshavallig

6)cotx * cot^2x=cotx(cosec^2x-1)
put t=cotx

8)take out a^2cos^2x as common from the denom,it will bcom sec^2x *1/a^2 in numerator
i=(1/a^2)sec^2x/(tan^2x+(b^2/a^2))
put t=tan^2x
apply the formula .in the formula A=b/a

varshavallig

9)e^-x =1/e^x
multiply and divide by 2 ,put t=e^2x +1 dt=2e^2x

10)e^-x=1/e^x
put e^x+1=t dt=e^x

13)i=tanx*tan^2x=tanx(sec^2x-1)
put t=tanx dt=sec^2x

!!!!!!!!!!!!!!!!!!!!!!!!!

uday_zingtudor

4=7)

sin2x=sin(5x-3x)

(sin5xcos3x-cos5xsin3x)sin5xsin3x = cot3x - cot5x

Integrating this, the answer is 1/3ln|sin3x| - 1/5ln|sin5x|

11)

x/rt(1-x²) + 2/rt(1-x²)Now integrating this, -rt(1-x²) +2sin^-1x

12) csc⁴x = csc²x(cot²x+1)

= csc²x + csc²xcot²x

Integrating this u get -cotx -1/3cot³x

~Cheerio!!!

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