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Question 373 of 373

Q.Evaluate : ∫0π/4sin⁡x+cos⁡x16+9sin⁡2x dx\int_{0}^{\pi/4} \dfrac{\sin x + \cos x}{16 + 9\sin 2x}\, dx OR Evaluate : ∫13(x2+3x+ex) dx\int_{1}^{3} (x^2 + 3x + e^x)\, dx as the limit of the sum.

Puducherry CbseCBSE Class XII Board 2018Subjective· 6mImportance★★★★★
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The value is 130ln⁡4=115ln⁡2\dfrac{1}{30}\ln4=\dfrac{1}{15}\ln2 (OR case: 623+e3−e\dfrac{62}{3}+e^3-e).

Concept. Substitution using (sin⁡x−cos⁡x)2=1−sin⁡2x(\sin x-\cos x)^2=1-\sin2x, then the standard form ∫dta2−t2=12aln⁡∣a+ta−t∣\displaystyle\int\frac{dt}{a^2-t^2}=\frac{1}{2a}\ln\left|\frac{a+t}{a-t}\right|.

Why this method. The numerator sin⁡x+cos⁡x\sin x+\cos x is exactly the derivative of t=sin⁡x−cos⁡xt=\sin x-\cos x.

Working. Let t=sin⁡x−cos⁡xt=\sin x-\cos x, dt=(cos⁡x+sin⁡x) dxdt=(\cos x+\sin x)\,dx, and sin⁡2x=1−t2\sin2x=1-t^2. Limits: x=0⇒t=−1x=0\Rightarrow t=-1; x=π4⇒t=0x=\tfrac\pi4\Rightarrow t=0.

I=∫−10dt16+9(1−t2)=∫−10dt25−9t2=19∫−10dt(53)2−t2.I=\int_{-1}^{0}\frac{dt}{16+9(1-t^2)}=\int_{-1}^{0}\frac{dt}{25-9t^2}=\frac19\int_{-1}^{0}\frac{dt}{\left(\tfrac53\right)^2-t^2}.

=19⋅12⋅53[ln⁡∣53+t53−t∣]−10=130[ln⁡1−ln⁡14]=130ln⁡4=115ln⁡2.=\frac19\cdot\frac{1}{2\cdot\tfrac53}\left[\ln\left|\frac{\tfrac53+t}{\tfrac53-t}\right|\right]_{-1}^{0}=\frac{1}{30}\left[\ln1-\ln\tfrac14\right]=\frac{1}{30}\ln4=\frac{1}{15}\ln2.

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