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Question: Evaluate : \(\int_{0}^{\pi/4}\frac{(\sin x + \cos x)}{(9 + 16\sin 2x)}\)dx -...

Evaluate : ∫0π/4(sin⁡x+cos⁡x)(9+16sin⁡2x)\int_{0}^{\pi/4}\frac{(\sin x + \cos x)}{(9 + 16\sin 2x)}dx -

A

120\frac{1}{20}ln 2

B

120\frac{1}{20}ln 4

C

120\frac{1}{20}ln 3

D

None of these

Answer

120\frac{1}{20}ln 3

Explanation

Solution

Let I =∫0π/4(sin⁡x+cos⁡x)(9+16sin⁡2x)\int_{0}^{\pi/4}\frac{(\sin x + \cos x)}{(9 + 16\sin 2x)} dx

Put sin x – cos x = t then (cos x + sin x) dx = d t and (sin x – cos x)2 = t2

Ž 1 – sin 2x = t2 Ž sin 2x = 1 – t2

When  x=0⇒t=–1x=π/4⇒t=0}\left. \ \begin{matrix} x = 0 \Rightarrow t = –1 \\ x = \pi/4 \Rightarrow t = 0 \end{matrix} \right\}

Then I = ∫–10dt9+16(1−t2)\int_{–1}^{0}\frac{dt}{9 + 16(1 - t^{2})}

= ∫–10dt25−16t2\int_{–1}^{0}\frac{dt}{25 - 16t^{2}}= 116\frac{1}{16} ∫−10dt(54)2−t2\int_{- 1}^{0}\frac{dt}{\left( \frac{5}{4} \right)^{2} - t^{2}}

= {ln⁡∣54+t54−t∣}−10\left\{ \ln\left| \frac{\frac{5}{4} + t}{\frac{5}{4} - t} \right| \right\}_{- 1}^{0}

= 140\frac { 1 } { 40 } {ln1−ln⁡19}\left\{ ln1 - \ln\frac{1}{9} \right\}

= 140\frac{1}{40}{0 + ln 9} = 140\frac{1}{40} 2 ln 3

= 120\frac{1}{20}ln 3.