The value of the integral
∫ (sin θ.sin 2θ(sin⁶θ+sin⁴θ+sin²θ)√(2sin⁴θ+3sin²θ+6))/(1-cos 2θ) dθ is :

Solution

Maths sir · 5 steps

Cancel the trig first, then push one t inside the root so the outside becomes its derivative.

1. Simplify the trig. Use 1-cos 2θ = 2 sin² θ and sin 2θ = 2 sin θ cos θ. The 2 sin² θ cancels, leaving cos θ (sin⁶ θ+sin⁴ θ+sin² θ)√(2 sin⁴ θ+3 sin² θ+6).
2. Substitute. Put t = sin θ, so dt = cos θ dθ and the integral becomes ∫ t²(t⁴+t²+1)√(2t⁴+3t²+6) dt. Move one t inside the root as t². This gives ∫ (t⁵+t³+t)√(2t⁶+3t⁴+6t²) dt.

   u = 2t⁶+3t⁴+6t², du = 12(t⁵+t³+t) dt

3. Integrate. The integral is 1/12∫ √u du = 1/12·2/3u^(3/2) = 1/18(2 sin⁶ θ+3 sin⁴ θ+6 sin² θ)^(3/2)+c.
4. Switch to cosine. No option matches this form, so put sin² θ = 1-x with x = cos² θ. Then 2(1-x)³+3(1-x)²+6(1-x) expands to (2+3+6)-(6+6+6)x+(6+3)x²-2x³. That is 11-18 cos² θ+9 cos⁴ θ-2 cos⁶ θ.
5. So. The integral is 1/18[11-18 cos² θ+9 cos⁴ θ-2 cos⁶ θ]^(3/2)+c, which is option (c).

The idea to keep. When a polynomial multiplies a square root, try to make the outside factor the derivative of what is inside the root. Pushing a power of t inside the root often does it.

Answer from the NTA final answer key. Solution by PrepWiser.