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Integral Calculus
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Aditya Arora
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24 Oct 2007 23:06:15 IST
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Well, as you go higher, it would be difficult to solve indefinite integrals. but definite integrals can be solved easily.
For eg:
[ ]
[ ] sin15x dx
[ ] sin15x dx[ ]
[ ] sin14x*sinxdx
[ ] sin14x*sinxdx[ ]
[ ] (1-cos2x)7 sinxdx
[ ] (1-cos2x)7 sinxdxPut t = cosx ......................... Eq. 1
differentiate both sides
dt = -sinxdx
So,
[ ]
[ ] -(1-t2)7 dt
[ ] -(1-t2)7 dtNow you will have to open the polynomial and then solve it
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25 Oct 2007 21:58:19 IST
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hints to evaluate
(sinx)^m*(cosx)^ndx where m and n are natural numbers
(i) if m is odd put cosx = t
(ii) if n is odd put sinx=t
(iii)if both m and n are odd you can substitute either sinx=t or cosx=t
(iv) if both m and n are even use trigonmetric identities
(v) if m and n are rational numbers and (m+n-2)/2 is a negative integer , then substitute cotx=t or tanx=t
i hope you are clear with my reply!!
(sinx)^m*(cosx)^ndx where m and n are natural numbers(i) if m is odd put cosx = t
(ii) if n is odd put sinx=t
(iii)if both m and n are odd you can substitute either sinx=t or cosx=t
(iv) if both m and n are even use trigonmetric identities
(v) if m and n are rational numbers and (m+n-2)/2 is a negative integer , then substitute cotx=t or tanx=t
i hope you are clear with my reply!!











