If f(x)=∫0sin⁡x cos−1⁡tdt+∫0cos⁡x sin−1⁡tdt , where 0<x<π2, then f(π/4) is

We have f(x)=∫0sin⁡x cos−1⁡tdt+∫0cos⁡x sin−1⁡tdt⇒f(x)=∫0sin⁡x π2−sin−1⁡tdt+∫0cos⁡x π2−cos−1⁡tdtf(x)=π2sin⁡x+π2cos⁡x−∫0sin⁡x sin−1⁡tdt−∫0cos⁡x cos−1⁡tdtf(x)=π2(sin⁡x+cos⁡x)−∫0sin⁡x sin−1⁡tdt+∫0cos⁡x cos−1⁡tdt Now, ∫0sin⁡x sin−1⁡tdt=∫0x θcos⁡θdθ, where t=sin⁡ θ∫0sin⁡x sin−1⁡tdt=[θsin⁡θ+cos⁡θ]0x=xsin⁡x+cos⁡x−1and ∫0cos⁡x cos−1⁡tdt=−∫0x αsin⁡αdα, where t=cos⁡α∫0cos⁡x cos−1⁡1dt=−[−αcos⁡α+sin⁡α]0x∫0cos⁡x cos−1⁡tdt=xcos⁡x−sin⁡xf(x)=π2(sin⁡x+cos⁡x) −[xsin ⁡x+cos ⁡x−1+xcos ⁡x−sin ⁡x]fπ4=π222−π42+12−1+π42−12 fπ4=π2−π22+1=π22+1

If f(x)=∫0sin⁡x cos−1⁡tdt+∫0cos⁡x sin−1⁡tdt , where 0<x<π2, then f(π/4) is

We have f(x)=∫0sin⁡x cos−1⁡tdt+∫0cos⁡x sin−1⁡tdt⇒f(x)=∫0sin⁡x π2−sin−1⁡tdt+∫0cos⁡x π2−cos−1⁡tdtf(x)=π2sin⁡x+π2cos⁡x−∫0sin⁡x sin−1⁡tdt−∫0cos⁡x cos−1⁡tdtf(x)=π2(sin⁡x+cos⁡x)−∫0sin⁡x sin−1⁡tdt+∫0cos⁡x cos−1⁡tdt Now, ∫0sin⁡x sin−1⁡tdt=∫0x θcos⁡θdθ, where t=sin⁡ θ∫0sin⁡x sin−1⁡tdt=[θsin⁡θ+cos⁡θ]0x=xsin⁡x+cos⁡x−1and ∫0cos⁡x cos−1⁡tdt=−∫0x αsin⁡αdα, where t=cos⁡α∫0cos⁡x cos−1⁡1dt=−[−αcos⁡α+sin⁡α]0x∫0cos⁡x cos−1⁡tdt=xcos⁡x−sin⁡xf(x)=π2(sin⁡x+cos⁡x) −[xsin ⁡x+cos ⁡x−1+xcos ⁡x−sin ⁡x]fπ4=π222−π42+12−1+π42−12 fπ4=π2−π22+1=π22+1

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If $f (x) = \int_{0}^{\sin x} \cos^{- 1} t d t + \int_{0}^{\cos x} \sin^{- 1} t d t$ , where $0 < x < \frac{π}{2},$ then $f (π / 4)$ is

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$\frac{π}{\sqrt{2}}$

$1 + \frac{π}{2 \sqrt{2}}$

none of these

answer is B.

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Detailed Solution

We have

$\begin{array}{l} f (x) = \int_{0}^{\sin x} \cos^{- 1} t d t + \int_{0}^{\cos x} \sin^{- 1} t d t \\ \Rightarrow f (x) = \int_{0}^{\sin x} (\frac{π}{2} - \sin^{- 1} t) d t + \int_{0}^{\cos x} (\frac{π}{2} - \cos^{- 1} t) d t \\ f (x) = \frac{π}{2} \sin x + \frac{π}{2} \cos x - \int_{0}^{\sin x} \sin^{- 1} t d t - \int_{0}^{\cos x} \cos^{- 1} t d t \\ f (x) = \frac{π}{2} (\sin x + \cos x) - [\int_{0}^{\sin x} \sin^{- 1} t d t + \int_{0}^{\cos x} \cos^{- 1} t d t] \end{array}$

Now, $\int_{0}^{\sin x} \sin^{- 1} t d t = \int_{0}^{x} θ \cos θ d θ,$ where $t = \sin θ$

$\int_{0}^{\sin x} \sin^{- 1} t d t = [θ \sin θ + \cos θ]_{0}^{x} = x \sin x + \cos x - 1$

and

$\begin{array}{l} \int_{0}^{\cos x} \cos^{- 1} t d t = - \int_{0}^{x} α \sin α d α, where t = \cos α \\ \int_{0}^{\cos x} \cos^{- 1} 1 d t = - [- α \cos α + \sin α]_{0}^{x} \\ \int_{0}^{\cos x} \cos^{- 1} t d t = x \cos x - \sin x \end{array}$

$f (x) = \frac{π}{2} (\sin x + \cos x)$

$- [x \sin x + \cos x - 1 + x \cos x - \sin x]$

$\begin{array}{l} f (\frac{π}{4}) = \frac{π}{2} (\frac{2}{\sqrt{2}}) - [\frac{π}{4 \sqrt{2}} + \frac{1}{\sqrt{2}} - 1 + \frac{π}{4 \sqrt{2}} - \frac{1}{\sqrt{2}}] \\ f (\frac{π}{4}) = \frac{π}{\sqrt{2}} - \frac{π}{2 \sqrt{2}} + 1 = \frac{π}{2 \sqrt{2}} + 1 \end{array}$