Notes On Integration Using Trigonometric Identities - CBSE Class 12 Maths
Integrals involving trigonometric functions, which cannot be solved using the direct approach. To solve such integrals, we use trigonometric identities. Ex: âˆ« cos4x dx I = âˆ« (cos2x)2 dx cos 2x = 2cos2x - 1 cos2x = (1+cos 2x)/2 I  = âˆ« ((1+cos 2x)/2)2 dx I  = âˆ« (1 + 2  cos 2x + cos22x)/4 dx    = 1/4 âˆ« (1 + 2 cos 2x + cos22x) dx    = 1/4 [âˆ« 1dx + 2 âˆ« cos 2x dx + âˆ« cos22x dx]    = 1/4[ âˆ« 1dx + 2 âˆ« cos 2x dx + âˆ« (1+cos 4x)/2 dx] ['.' cos22x = (1+cos 4x)/2]    = 1/4[ âˆ« 1dx + 2 âˆ« cos 2x dx + 1/2[âˆ« 1 dx + âˆ« cos 4x dx]]    = 1/4 âˆ« 1dx + 1/2 âˆ« cos 2x dx + âˆ« 1 dx + 1/8 âˆ« cos 4x dx    = 1/4 x + 1/2 .(sin 2x)/2 + 1/8 . x + 1/8 (sin 4x)/4 + C    = 1/4 .x + 1/4 sin 2x + 1/8 . x + 1/32 . sin 4x + C    = 3/8 . x + 1/4 sin 2x + 1/32 sin 4x + C    = 1/32[12x + 8 sin 2x + sin 4x] + C âˆ« cos4x dx = 1/32[12x + 8 sin 2x + sin 4x] + C

#### Summary

Integrals involving trigonometric functions, which cannot be solved using the direct approach. To solve such integrals, we use trigonometric identities. Ex: âˆ« cos4x dx I = âˆ« (cos2x)2 dx cos 2x = 2cos2x - 1 cos2x = (1+cos 2x)/2 I  = âˆ« ((1+cos 2x)/2)2 dx I  = âˆ« (1 + 2  cos 2x + cos22x)/4 dx    = 1/4 âˆ« (1 + 2 cos 2x + cos22x) dx    = 1/4 [âˆ« 1dx + 2 âˆ« cos 2x dx + âˆ« cos22x dx]    = 1/4[ âˆ« 1dx + 2 âˆ« cos 2x dx + âˆ« (1+cos 4x)/2 dx] ['.' cos22x = (1+cos 4x)/2]    = 1/4[ âˆ« 1dx + 2 âˆ« cos 2x dx + 1/2[âˆ« 1 dx + âˆ« cos 4x dx]]    = 1/4 âˆ« 1dx + 1/2 âˆ« cos 2x dx + âˆ« 1 dx + 1/8 âˆ« cos 4x dx    = 1/4 x + 1/2 .(sin 2x)/2 + 1/8 . x + 1/8 (sin 4x)/4 + C    = 1/4 .x + 1/4 sin 2x + 1/8 . x + 1/32 . sin 4x + C    = 3/8 . x + 1/4 sin 2x + 1/32 sin 4x + C    = 1/32[12x + 8 sin 2x + sin 4x] + C âˆ« cos4x dx = 1/32[12x + 8 sin 2x + sin 4x] + C

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