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Integrals of the form int sinm xcosn x dx in learning the technique of substitution we saw the integral int sin xcos x dx in example 614.
Integral sinm x cosn x dx formula. Then use the change of variable u sinx. The reduction formula can be derived using any of the common methods of integration like integration by substitution integration by parts integration by trigonometric substitution integration by partial fractions etcthe main idea is to express an integral involving an integer parameter eg. We start by using double angle formulas z sin2xcos4xdx z secxcosx2 cos2xdx z 1 4 sin22xcos2xdx 1 8 z sin22x1cos2xdx 1 16 z 1 cos4x1cos2xdx 1 16 z 6 1 cos4xcos2x cos4xcos2xdx and using 4 we we can compute.
As an application of the previous example we compute the integral z sin2xcos4xdx. The integration was not difficult and one could easily evaluate the indefinite integral by letting usin x or by letting u cos x. Reduction formula of sinmxcosnx reduction formula for sinmx cosnx reduction formula of sinnxcosmx.
Power of a function represented by i n in terms of an. Solution we could evaluate this integral using the reduction formula for equation 567 together with example 3 as in exercise 33 in section 56 but a better. 598 integration techniques if the exponent of cosine is odd split off one cosx and use the identity cos2x 1 sin2x to rewrite the remaining even power of cosine in terms of sine.
Example z tanxdx lnjsecxjc proof z tanxdx sinx cosx dx using the substitution u cosx we get du sinx giving us that the above integral is. Reduction formula is very useful in solving integration problem. Powers of tangent can be reduced using the formula tan2 x sec2 x 1 example z tan0 xdx z 1dx xc.
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