If \int \operatornamecosec^5 x d x=\alpha \cot x \operatornamecosec x(\operatornamecosec^2 x+\frac32)+\beta \log _x|\tan \fracx2|+C where \alpha, \beta \in \mathbbR and C is the constant of integration, then the value of 8(\alpha+\beta) equals \underline\hspace2cm .

Indefinite Integrals PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q52 with Solution

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Correct answer: 1

We need to evaluate $I=\int \csc^5 x\,dx$ and compare it with $I=\alpha \cot x\csc x\left(\csc^2 x+\frac32\right)+\beta \log\left|\tan\frac x2\right|+C.$
Use the standard reduction formula for $I_n=\int \csc^n x\,dx.$ For odd powers, $I_n=-\frac{\csc^{n-2}x\cot x}{n-1}+\frac{n-2}{n-1}I_{n-2}.$

For $n=5$ , $I_5=-\frac{\csc^3x\cot x}{4}+\frac34 I_3.$

Now compute $I_3$ : $I_3=\int \csc^3x\,dx=-\frac12\csc x\cot x+\frac12\int \csc x\,dx.$ And $\int \csc x\,dx=\log\left|\tan\frac x2\right|+C.$ Hence, $I_3=-\frac12\csc x\cot x+\frac12\log\left|\tan\frac x2\right|+C.$
Substitute into $I_5$ : $I_5=-\frac14\csc^3x\cot x+\frac34\left(-\frac12\csc x\cot x+\frac12\log\left|\tan\frac x2\right|\right)+C.$ So, $I_5=-\frac14\csc^3x\cot x-\frac38\csc x\cot x+\frac38\log\left|\tan\frac x2\right|+C.$
Factor the first two terms:

= -\frac14\csc x\cot x\left(\csc^2x+\frac32\right).$$ Thus, $$\int \csc^5x\,dx=-\frac14\cot x\csc x\left(\csc^2x+\frac32\right)+\frac38\log\left|\tan\frac x2\right|+C.$$ 6. Comparing with the given form, $$\alpha=-\frac14,\qquad \beta=\frac38.$$ Therefore, $$\alpha+\beta=-\frac14+\frac38=\frac18.$$ So, $$8(\alpha+\beta)=8\cdot \frac18=1.$$ 7. Final answer: $$\boxed{1}$$ The derived answer matches the stored correct answer.

View written solutionFree

Correct answer: 1

We need to evaluate $I=\int \csc^5 x\,dx$ and compare it with $I=\alpha \cot x\csc x\left(\csc^2 x+\frac32\right)+\beta \log\left|\tan\frac x2\right|+C.$
Use the standard reduction formula for $I_n=\int \csc^n x\,dx.$ For odd powers, $I_n=-\frac{\csc^{n-2}x\cot x}{n-1}+\frac{n-2}{n-1}I_{n-2}.$

For $n=5$ , $I_5=-\frac{\csc^3x\cot x}{4}+\frac34 I_3.$

Now compute $I_3$ : $I_3=\int \csc^3x\,dx=-\frac12\csc x\cot x+\frac12\int \csc x\,dx.$ And $\int \csc x\,dx=\log\left|\tan\frac x2\right|+C.$ Hence, $I_3=-\frac12\csc x\cot x+\frac12\log\left|\tan\frac x2\right|+C.$
Substitute into $I_5$ : $I_5=-\frac14\csc^3x\cot x+\frac34\left(-\frac12\csc x\cot x+\frac12\log\left|\tan\frac x2\right|\right)+C.$ So, $I_5=-\frac14\csc^3x\cot x-\frac38\csc x\cot x+\frac38\log\left|\tan\frac x2\right|+C.$
Factor the first two terms:

= -\frac14\csc x\cot x\left(\csc^2x+\frac32\right).$$ Thus, $$\int \csc^5x\,dx=-\frac14\cot x\csc x\left(\csc^2x+\frac32\right)+\frac38\log\left|\tan\frac x2\right|+C.$$ 6. Comparing with the given form, $$\alpha=-\frac14,\qquad \beta=\frac38.$$ Therefore, $$\alpha+\beta=-\frac14+\frac38=\frac18.$$ So, $$8(\alpha+\beta)=8\cdot \frac18=1.$$ 7. Final answer: $$\boxed{1}$$ The derived answer matches the stored correct answer.