Let $$x2$$≠nπ−$$1$$,n∈N . If ∫$$x2sin$$⁡$$x2+1$$−$$sin$$⁡$$2x2+12sin$$⁡$$x2+1+$$$$sin$$⁡$$2x2+1dx=ln$$⁡$$f(x)+C$$ then $$sec$$−$$1$$⁡$$f(1)$$ equals

Let $$x2+1=t$$⇒$$xdx=dt2Given$$ integral $$=12$$∫$$2sin$$⁡t−$$2sin$$⁡tcos⁡$$t2sin$$⁡$$t+2sin$$⁡tcos⁡$$tdt=12$$∫$$tan$$⁡$$t2dt=ln$$⁡$$sec$$⁡$$t2+C$$∴$$f(x)=$$$$sec$$⁡$$x2+12sec$$−$$1$$⁡$$f(1)=1$$

Methods of integration

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Question

Let $x^{2} \neq nπ - 1, n \in N$ . If $\int x \sqrt{\frac{2 \sin (x^{2} + 1) - \sin 2 (x^{2} + 1)}{2 \sin (x^{2} + 1) + \sin 2 (x^{2} + 1)}} dx = \ln f (x) + C$ then $\sec^{- 1} f (1)$ equals

Moderate

Solution

Let $x^{2} + 1 = t \Rightarrow xdx = \frac{dt}{2}$
Given integral $= \frac{1}{2} \int \sqrt{\frac{2 \sin t - 2 \sin tcos t}{2 \sin t + 2 \sin tcos t}} dt$
$\begin{array}{l} = \frac{1}{2} \int \tan \frac{t}{2} dt = \ln |\sec \frac{t}{2}| + C \\ ∴ f (x) = \sec (\frac{x^{2} + 1}{2}) \\ \sec^{- 1} f (1) = 1 \end{array}$

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Methods of integration

Unlock the full solution & master the concept.

Let x2≠nπ−1,n∈N . If ∫x2sin⁡x2+1−sin⁡2x2+12sin⁡x2+1+sin⁡2x2+1dx=ln⁡f(x)+C then sec−1⁡f(1) equals

Let x2+1=t⇒xdx=dt2Given integral =12∫2sin⁡t−2sin⁡tcos⁡t2sin⁡t+2sin⁡tcos⁡tdt=12∫tan⁡t2dt=ln⁡sec⁡t2+C∴f(x)=sec⁡x2+12sec−1⁡f(1)=1

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Let $x^{2} \neq nπ - 1, n \in N$ . If $\int x \sqrt{\frac{2 \sin (x^{2} + 1) - \sin 2 (x^{2} + 1)}{2 \sin (x^{2} + 1) + \sin 2 (x^{2} + 1)}} dx = \ln f (x) + C$ then $\sec^{- 1} f (1)$ equals