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

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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



\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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