IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

Werewolf

What is the general approach to solve the integrals of the type

sin^mxcosⁿxdx , when the power of sinx and cosx are both positive even integers.

konichiwa2x

Trigonometric formulas.

siddharthsaxena

You have to use the following formulae :-

cos²x = (1 + cos2x) / 2

sin²x = (1-cos2x) / 2

Since m , n are even positive , you will get the integral in terms of above formulae

Aftr dat its easy dude

Hope its useful

Cheers!!!!!!!!!

karthik2007

U can use gamma function if its a definite integral

ananth_patri

there are many formulae for diff conditions...but just follow the logic and solve using the method posted by sidharthsaxena......its always usefull....better than gettin confused with all those formulae....

Werewolf

But what if the powers are big , the integral is easy when one of the power is odd but what if they both are even and large powers like m=10 n=8 or any other .

chimanshu_007

sorry gave the wrong answer

i have learnt

integral sin^mx cosⁿx = sin^m+1x cosⁿ⁺¹x / m + n + (n-1/m+n) integral sin^mx.cosⁿ⁺² dx

u can solve dis

Werewolf

Actually once I found a friend of mine using a complex number approach in these kind of integrals involving even powers only and I wanted to know that.If anyone knows that method please share it with me.

In the earlier method chimanshu told we again face the same kind of integral after applying the integration by parts formula so it becomes a bit lengthy if powers are bit large, that complex method technique had very short steps leading to the answer but I unfortunately forgot that method.So it will be helpful if anyone could remind me that method.

siddharthsaxena

Yes, thats using GAMMA function . will give u details in some time

konichiwa2x

chimanshu, it wont be very advisable to learn such formulas. Three is no limit to the number of formulas if u start doing such stuff.

werewolf, you are right. expressing sin

and cos

in terms of complex variables is one way of doing it. But mostly, trigonmetric formulas the best even if powers are large.

chimanshu_007

well i know 4-5 formulas for this kind only but i have learnt only one....it works for me....dunno bout others......

layman

If m is odd put cosx=t,

If n is odd put sinx=t,

If both are odd any of the above substitution,

If both are even convert them into a combination of sin(rx) and cos(rx) and then solve.

eistien

cos²x = (1 + cos2x) / 2

sin²x = (1-cos2x) / 2

IF THE POWERS ARE EVEN !!!

this is in addition to what layman has said!!!

layman

One more good method if powers are even:

cos x=1/2(z+1/z)

sin x=1/2i(z-1/z)

Then expand binomially then finally put:

z^n +1/z^n =2cos(nx)

and solve.

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