CK-12-Calculus

7.4. Trigonometric Integrals http://www.ck12.org

=−cosx−

∫ [

−u^2 sinxsindux

]

=−cosx+

∫

u^2 du

=−cosx+u

3

3 +C

=−cosx+^13 cos^3 x+C.

Ifmandnare both positive integers, then an integral of the form

∫

sinmxcosnxdx

can be evaluated by one of the procedures shown in the table below, depending on whethermandnare odd or even.

TABLE7.2:

∫sinmxcosnxdx Procedure Identities

nodd Letu=sinx cos^2 x= 1 −sin^2 x

modd Letu=cosx sin^2 x= 1 −cos^2 x

nandmeven Use identities to reduce powers sin^2 x= ( 1 / 2 )( 1 −cos 2x)

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

Example 4:
Evaluate∫sin^3 xcos^4 xdx.
Solution:
Here,mis odd. So according to the second procedure in the table above, letu=cosx,sodu=−sinx.Substituting,

∫

sin^3 xcos^4 xdx=

∫

u^4 sin^3 xsin−^1 xdu

=−

∫

u^4 sin^2 xdu.

Referring to the table again, we can now substitute sin^2 x= 1 −cos^2 xin the integral:

=−

∫

u^4 ( 1 −cos^2 x)du

=−

∫

u^4 ( 1 −u^2 )du

=

∫

(−u^4 +u^6 )du

=− 51 u^5 +^17 u^7 +C

=−^15 cos^5 x+^17 cos^7 x+C.