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Integral Calculus
can u plz integrate these
1)sec5xtanx ans 1/5(sec5x)
2)(cos2x)/(sinx +cosx)2 = ans log lsinx+cosxl
3)(xe-1 + ex-1 )/(xe + ex ) ans 1/e{logIxe +exl}
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/(ex + e-x )2 ans -1/(2(1+e2x ))
10) 1/(ex +1)(1+e-x ) ans -1/(1+ex )
11)(x+2)/
(1-x2 ) ans -
(1-x2 ) + 2sin-x
12)cosec4x ans -cotx -(cot3x)/3
13) tan3x ans -loglsecxl +(tan2 x)/2
Comments (4)
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!!!












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
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