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Indefinite Integrals question

2019 · 10 Jan · Shift 1 · Q26
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  5. /2019 · 10 Jan · Shift 1 · Q26

Indefinite Integrals question

2019 · 10 Jan · Shift 1 · Q26

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
Let n ≥\ge≥ 2 be a natural number and 0<θ<π2.0 \lt \theta \lt {\pi \over 2}.0<θ<2π​. Then ∫(sin⁡nθ−sin⁡θ)1/ncos⁡θsin⁡n+1θ dθ\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta∫sinn+1θ(sinnθ−sinθ)1/ncosθ​dθ is equal to - (where C is a constant of integration)
  1. A
    nn2−1(1+1sin⁡n−1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 + {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + Cn2−1n​(1+sinn−1θ1​)nn+1​+C
  2. B
    nn2−1(1−1sin⁡n+1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n + 1}}\theta }}} \right)^{{{n + 1} \over n}}} + Cn2−1n​(1−sinn+1θ1​)nn+1​+C
  3. C
    nn2−1(1−1sin⁡n−1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + Cn2−1n​(1−sinn−1θ1​)nn+1​+C
  4. D
    nn2+1(1−1sin⁡n−1θ)n+1n+C{n \over {{n^2} + 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + Cn2+1n​(1−sinn−1θ1​)nn+1​+C
View written solutionFree

Correct answer: C

  1. Given integral

We need to evaluate I=∫(sin⁡nθ−sin⁡θ)1/ncos⁡θsin⁡n+1θ dθ,I=\int \frac{\left(\sin^n\theta-\sin\theta\right)^{1/n}\cos\theta}{\sin^{n+1}\theta}\,d\theta,I=∫sinn+1θ(sinnθ−sinθ)1/ncosθ​dθ, where n≥2n\ge 2n≥2 and 0<θ<π20<\theta<\frac\pi20<θ<2π​.

  1. Simplify the expression inside the root

Factor sin⁡θ\sin\thetasinθ from sin⁡nθ−sin⁡θ\sin^n\theta-\sin\thetasinnθ−sinθ: sin⁡nθ−sin⁡θ=sin⁡θ(sin⁡n−1θ−1).\sin^n\theta-\sin\theta=\sin\theta\left(\sin^{n-1}\theta-1\right).sinnθ−sinθ=sinθ(sinn−1θ−1).

A better way for substitution is to rewrite directly: sin⁡nθ−sin⁡θ=sin⁡θ(sin⁡n−1θ−1).\sin^n\theta-\sin\theta=\sin\theta\big(\sin^{n-1}\theta-1\big).sinnθ−sinθ=sinθ(sinn−1θ−1).

Since 0<θ<π20<\theta<\frac\pi20<θ<2π​, we have sin⁡θ>0\sin\theta>0sinθ>0. So powers are well-defined in the intended real form.

Now write

=\left[\sin\theta\left(\sin^{n-1}\theta-1\right)\right]^{1/n}.$$ Then $$I=\int \frac{\left[\sin\theta(\sin^{n-1}\theta-1)\right]^{1/n}\cos\theta}{\sin^{n+1}\theta}\,d\theta.$$ This is messy as-is, so let us express the factor in terms of $\sin^{-(n-1)}\theta$. Observe: $$\sin^n\theta-\sin\theta=\sin\theta\left(\sin^{n-1}\theta-1\right) = -\sin\theta\left(1-\sin^{n-1}\theta\right).$$ Divide by $\sin^{n+1}\theta$ and combine powers: $$\frac{(\sin^n\theta-\sin\theta)^{1/n}}{\sin^{n+1}\theta} =\frac{\left[\sin\theta(\sin^{n-1}\theta-1)\right]^{1/n}}{\sin^{n+1}\theta} =\frac{(\sin^{n-1}\theta-1)^{1/n}}{\sin^{n+1-1/n}\theta}.$$ Now note that $$\sin^{n-1}\theta-1=\sin^{n-1}\theta\left(1-\frac1{\sin^{n-1}\theta}\right).$$ Hence $$\left(\sin^{n-1}\theta-1\right)^{1/n} =\sin^{\frac{n-1}{n}}\theta\left(1-\frac1{\sin^{n-1}\theta}\right)^{1/n}.$$ Therefore, $$\frac{(\sin^{n-1}\theta-1)^{1/n}}{\sin^{n+1-1/n}\theta} =\frac{\sin^{(n-1)/n}\theta}{\sin^{n+1-1/n}\theta} \left(1-\frac1{\sin^{n-1}\theta}\right)^{1/n}.$$ The exponent of $\sin\theta$ in the denominator becomes $$n+1-\frac1n-\frac{n-1}{n}=n,$$ so $$I=\int \frac{\cos\theta}{\sin^n\theta}\left(1-\frac1{\sin^{n-1}\theta}\right)^{1/n}d\theta.$$ --- 3. **Substitution** Let $$u=1-\frac1{\sin^{n-1}\theta}=1-\sin^{-(n-1)}\theta.$$ Differentiate: $$\frac{du}{d\theta}=-( -(n-1)\sin^{-n}\theta\cos\theta) =(n-1)\frac{\cos\theta}{\sin^n\theta}.$$ So, $$du=(n-1)\frac{\cos\theta}{\sin^n\theta}\,d\theta,$$ which gives $$\frac{\cos\theta}{\sin^n\theta}\,d\theta=\frac{du}{n-1}.$$ Thus the integral becomes $$I=\frac1{n-1}\int u^{1/n}\,du.$$ --- 4. **Integrate** $$\int u^{1/n}\,du=\frac{u^{\frac{n+1}{n}}}{\frac{n+1}{n}}=\frac{n}{n+1}u^{\frac{n+1}{n}}.$$ Hence $$I=\frac1{n-1}\cdot \frac{n}{n+1}u^{\frac{n+1}{n}}+C =\frac{n}{n^2-1}u^{\frac{n+1}{n}}+C.$$ Substituting back $u$: $$I=\frac{n}{n^2-1}\left(1-\frac1{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}}+C.$$ --- 5. **Match with options** This is exactly **Option C**: $$\boxed{\frac{n}{n^2-1}\left(1-\frac1{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}}+C}.$$ --- 6. **Verification with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.
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