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Question: Let $n \geq 2$ be a natural number and $0 < \theta < \pi/2$. Then $\int \frac{(\sin^n \theta - \sin ...

Let n≥2n \geq 2 be a natural number and 0<θ<π/20 < \theta < \pi/2. Then ∫(sin⁡nθ−sin⁡θ)1ncos⁡θsin⁡n+1θdθ\int \frac{(\sin^n \theta - \sin \theta)^{\frac{1}{n}} \cos \theta}{\sin^{n+1} \theta} d\theta is equal to: (2019)

(where C is a constant of integration)

A

nn2+1(1−1sin⁡n−1θ)n+1n+C\frac{n}{n^2+1} (1-\frac{1}{\sin^{n-1}\theta})^{\frac{n+1}{n}} + C

B

nn2−1(1−1sin⁡n−1θ)n+1n+C\frac{n}{n^2-1} (1-\frac{1}{\sin^{n-1}\theta})^{\frac{n+1}{n}} + C

C

nn2−1(1−1sin⁡n−1θ)n+1n+C\frac{n}{n^2-1} (1-\frac{1}{\sin^{n-1}\theta})^{\frac{n+1}{n}} + C

D

nn2−1(1+1sin⁡n−1θ)n+1n+C\frac{n}{n^2-1} (1+\frac{1}{\sin^{n-1}\theta})^{\frac{n+1}{n}} + C

Answer

nn2−1(1−1sin⁡n−1θ)n+1n+C\frac{n}{n^2-1} (1-\frac{1}{\sin^{n-1}\theta})^{\frac{n+1}{n}} + C

Explanation

Solution

The given integral is I=∫(sin⁡nθ−sin⁡θ)1ncos⁡θsin⁡n+1θdθI = \int \frac{(\sin^n \theta - \sin \theta)^{\frac{1}{n}} \cos \theta}{\sin^{n+1} \theta} d\theta.

Step 1: Simplify the term inside the parenthesis in the numerator.

Factor out sin⁡nθ\sin^n \theta from (sin⁡nθ−sin⁡θ)(\sin^n \theta - \sin \theta): sin⁡nθ−sin⁡θ=sin⁡nθ(1−sin⁡θsin⁡nθ)=sin⁡nθ(1−1sin⁡n−1θ)\sin^n \theta - \sin \theta = \sin^n \theta \left(1 - \frac{\sin \theta}{\sin^n \theta}\right) = \sin^n \theta \left(1 - \frac{1}{\sin^{n-1} \theta}\right).

Now, raise this to the power of 1n\frac{1}{n}: (sin⁡nθ−sin⁡θ)1n=(sin⁡nθ(1−1sin⁡n−1θ))1n(\sin^n \theta - \sin \theta)^{\frac{1}{n}} = \left(\sin^n \theta \left(1 - \frac{1}{\sin^{n-1} \theta}\right)\right)^{\frac{1}{n}} =(sin⁡nθ)1n(1−1sin⁡n−1θ)1n= (\sin^n \theta)^{\frac{1}{n}} \left(1 - \frac{1}{\sin^{n-1} \theta}\right)^{\frac{1}{n}} =sin⁡θ(1−1sin⁡n−1θ)1n= \sin \theta \left(1 - \frac{1}{\sin^{n-1} \theta}\right)^{\frac{1}{n}}.

Step 2: Substitute this simplified term back into the integral. I=∫sin⁡θ(1−1sin⁡n−1θ)1ncos⁡θsin⁡n+1θdθI = \int \frac{\sin \theta \left(1 - \frac{1}{\sin^{n-1} \theta}\right)^{\frac{1}{n}} \cos \theta}{\sin^{n+1} \theta} d\theta I=∫(1−1sin⁡n−1θ)1ncos⁡θsin⁡nθdθI = \int \frac{\left(1 - \frac{1}{\sin^{n-1} \theta}\right)^{\frac{1}{n}} \cos \theta}{\sin^n \theta} d\theta.

Step 3: Use substitution. Let t=1−1sin⁡n−1θt = 1 - \frac{1}{\sin^{n-1} \theta}. This can be written as t=1−(sin⁡θ)−(n−1)t = 1 - (\sin \theta)^{-(n-1)}. Now, differentiate tt with respect to θ\theta: dtdθ=0−(−(n−1))(sin⁡θ)−(n−1)−1(cos⁡θ)\frac{dt}{d\theta} = 0 - (-(n-1)) (\sin \theta)^{-(n-1)-1} (\cos \theta) dtdθ=(n−1)(sin⁡θ)−ncos⁡θ\frac{dt}{d\theta} = (n-1) (\sin \theta)^{-n} \cos \theta dtdθ=(n−1)cos⁡θsin⁡nθ\frac{dt}{d\theta} = (n-1) \frac{\cos \theta}{\sin^n \theta}.

From this, we can express cos⁡θsin⁡nθdθ\frac{\cos \theta}{\sin^n \theta} d\theta in terms of dtdt: cos⁡θsin⁡nθdθ=1n−1dt\frac{\cos \theta}{\sin^n \theta} d\theta = \frac{1}{n-1} dt.

Step 4: Substitute tt and dtdt into the integral. I=∫t1n(1n−1)dtI = \int t^{\frac{1}{n}} \left(\frac{1}{n-1}\right) dt I=1n−1∫t1ndtI = \frac{1}{n-1} \int t^{\frac{1}{n}} dt.

Step 5: Integrate with respect to tt. I=1n−1(t1n+11n+1)+CI = \frac{1}{n-1} \left(\frac{t^{\frac{1}{n}+1}}{\frac{1}{n}+1}\right) + C I=1n−1(tn+1nn+1n)+CI = \frac{1}{n-1} \left(\frac{t^{\frac{n+1}{n}}}{\frac{n+1}{n}}\right) + C I=1n−1(nn+1tn+1n)+CI = \frac{1}{n-1} \left(\frac{n}{n+1} t^{\frac{n+1}{n}}\right) + C I=n(n−1)(n+1)tn+1n+CI = \frac{n}{(n-1)(n+1)} t^{\frac{n+1}{n}} + C I=nn2−1tn+1n+CI = \frac{n}{n^2-1} t^{\frac{n+1}{n}} + C.

Step 6: Substitute back t=1−1sin⁡n−1θt = 1 - \frac{1}{\sin^{n-1} \theta}. I=nn2−1(1−1sin⁡n−1θ)n+1n+CI = \frac{n}{n^2-1} \left(1 - \frac{1}{\sin^{n-1} \theta}\right)^{\frac{n+1}{n}} + C.

Comparing this result with the given options, option (B) and (C) are identical and match our derived solution.