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  • Provide Solution For R D Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18 Point 25 Question 29 Maths Textbook Solution

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 29 Maths Textbook Solution.

Answers (1)

Answer:\frac{-1}{2}\cos ecx\cot \, x+\frac{1}{2}log|\tan \frac{x}{2}|+c

Hint: \cos \: e\: c^{3}x=\cos ec^{2}x\: \cos \: e\: cx

Given:\int \cos \: ec^{3}xdx

Solution:

Let

\begin{aligned} &I=\int \operatorname{cosec}^{3} d x \\ &=\int \operatorname{cosec}^{2} x \cdot \operatorname{cosec} x d x \\ &=\int \operatorname{cosec}^{2} x \cdot \sqrt{1+\cot ^{2} x} \end{aligned}

let\: \cot \: x=t

\begin{aligned} &I=-\int \sqrt{1+t^{2}} d t \\ &=-\frac{t}{2} \sqrt{1+t^{2}}-\frac{1^{2}}{2} \log \left|t+\sqrt{1+t^{2}}\right|+C \end{aligned}

Substituting the value of t in equation (1)

\begin{aligned} &=-\frac{\cot x}{2} \cdot \cos e c x-\frac{1}{2} \log |\cot x+\cos e c x|+C \\ &=-\frac{1}{2} \cos e \cot x-\frac{1}{2} \log \left|\frac{\cos x}{\sin x}+\frac{1}{\sin x}\right|+C \\ &=-\frac{1}{2} \cos e c x \cot x-\frac{1}{2} \log \left|\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}\right|+C \\ &=-\frac{1}{2} \cos e \operatorname{cx} \cot x-\frac{1}{2} \log \left|\cot \frac{x}{2}\right|+C \\ &=-\frac{1}{2} \cos e c x \cot x+\frac{1}{2} \log \left|\tan \frac{x}{2}\right|+C\left[\because \log \left|\cot \frac{x}{2}\right|=\log \left|\frac{1}{\tan \frac{x}{2}}\right| \Rightarrow-\log \left|\tan \frac{x}{2}\right|\right] \end{aligned}

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