IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

sandeepramesh

put sinx = t :)

Radon222

that was not worth to be called "the demon"

pink_ele

Cos³x+cos⁵x/(sin²x+sin⁴x)

Here put sinx =t

Den u get .finally

1-t²+(1-t²)2/(t²+t⁴)

Simplifying further

Its 2-3t²+t⁴/(t²(1+t²)

Now

Break d integral accordingly n u get

2/t² -5/(1+t²) +t²/(1+t²)

Now

Its

2/t² -5/(1+t²) +1 -1/(1+t2)

Its easy to solve now I suppose

N give=-2/sinx -6tan-1(sinx) +sinx

hope it helped !

_{sorry dear i hav not put in tegral sign hope it still worked}

punnima

@pink_ele

"Here put sinx =t

Den u get .finally

1-t²+(1-t²)2/(t²+t⁴)" ?????????????????????????

i did not get it.!!!

pink_ele

thats

1-t²+(1-t²)²/(t²+t⁴)

dear

m sry

layman

$\int \frac {\cos ^{3}{x} + \cos ^{5}{x}}{\sin ^{2}{x} + \sin ^{4}{x}}\,dx$

= $\int \frac {\cos ^{3}{x}\cdot(1 + \cos ^{2}{x})}{\sin ^{2}{x} + \sin ^{4}{x}}\,dx$

Put $\sin{x} = t$ obviously, $dt = \cos{x}\cdot dx$

= $\int \frac {(1 - t^{2})\cdot (2 - t^{2})}{t^{2}\cdot (1 + t^{2})}\,dt$

= $\int \frac {t^{4} - 3\cdot t^{2} + 2}{t^{2}\cdot (1 + t^{2})}\,dt$

= $\int 1 + (\frac {2 - 4\cdot t^{2}}{t^{2}\cdot (1 + t^{2})})\,dt$

Doing partial fractions we get,

= $\int 1 + \frac {2}{t^{2}} - \frac {6}{1 + t^{2}}\,dt$

= $t - \frac {2}{t} - 6\cdot \tan^{ - 1} {t} + C$

= $\sin{x} - 2\cdot cosec x - 6\cdot \tan^{ - 1} {\sin{x}} + C$

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