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: ∫ cos4x dx

I = ∫ (cos2x)2 dx

cos 2x = 2cos2x - 1

cos2x = (1+cos 2x)/2

I  = ∫ ((1+cos 2x)/2)2 dx

I  = ∫ (1 + 2  cos 2x + cos22x)/4 dx

   = 1/4 ∫ (1 + 2 cos 2x + cos22x) dx

   = 1/4 [∫ 1dx + 2 ∫ cos 2x dx + ∫ cos22x dx]

   = 1/4[ ∫ 1dx + 2 ∫ cos 2x dx + ∫ (1+cos 4x)/2 dx] ['.' cos22x = (1+cos 4x)/2]

   = 1/4[ ∫ 1dx + 2 ∫ cos 2x dx + 1/2[∫ 1 dx + ∫ cos 4x dx]]

   = 1/4 ∫ 1dx + 1/2 ∫ cos 2x dx + ∫ 1 dx + 1/8 ∫ 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

∫ 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: ∫ cos4x dx

I = ∫ (cos2x)2 dx

cos 2x = 2cos2x - 1

cos2x = (1+cos 2x)/2

I  = ∫ ((1+cos 2x)/2)2 dx

I  = ∫ (1 + 2  cos 2x + cos22x)/4 dx

   = 1/4 ∫ (1 + 2 cos 2x + cos22x) dx

   = 1/4 [∫ 1dx + 2 ∫ cos 2x dx + ∫ cos22x dx]

   = 1/4[ ∫ 1dx + 2 ∫ cos 2x dx + ∫ (1+cos 4x)/2 dx] ['.' cos22x = (1+cos 4x)/2]

   = 1/4[ ∫ 1dx + 2 ∫ cos 2x dx + 1/2[∫ 1 dx + ∫ cos 4x dx]]

   = 1/4 ∫ 1dx + 1/2 ∫ cos 2x dx + ∫ 1 dx + 1/8 ∫ 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

∫ cos4x dx = 1/32[12x + 8 sin 2x + sin 4x] + C

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