Integral Calculus I Exercise 2.9 - 12th Text Book Solutions

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Important Formulas in Integral Calculus I

Text Book Solutions for Integral Calculus I Exercise 2.1

Text Book Solutions for Integral Calculus I Exercise 2.2

Text Book Solutions for Integral Calculus I Exercise 2.3

Text Book Solutions for Integral Calculus I Exercise 2.4

Text Book Solutions for Integral Calculus I Exercise 2.5

Text Book Solutions for Integral Calculus I Exercise 2.6

Text Book Solutions for Integral Calculus I Exercise 2.7

Text Book Solutions for Integral Calculus I Exercise 2.8

Text Book Solutions for Integral Calculus I Exercise 2.10

Text Book Solutions for Integral Calculus I Exercise 2.11

Text Book Solutions for Integral Calculus I Exercise 2.12 (MCQ)

Exercise 2.9

Evaluate the following using properties of definite integrals:

$1. \displaystyle \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}x^{3}\:cos^{3}x \: dx$

$I= \displaystyle \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}x^{3}\:cos^{3}x \: dx\\$
$Let \: f(x) =x^{3}\:cos^{3}x \: dx\\$
$f(-x) =(-x)^{3}\:cos(-x)^{3} \: dx\\$
$=-\left (x^{3}\:cos^{3}x \right )\: dx\\$
$f(-x)=-f(x)\\$
$f(x)\: is\: an \: odd \: function.\\$
$\therefore I=0$

$2. \displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}sin^{2}\theta \: d\theta$

$I= \displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}sin^{2}\theta \: d\theta \\$
$f(\theta)=sin^{2}\theta\\$
$f(\theta)=(sin\theta)^{2}\\$
$f(-\theta)=(-sin\theta)^{2}\\$
$f(-\theta)=sin^{2}\theta\\$
$f(-\theta)=f(\theta)\\$
$f(\theta)\: is \: an \: even \: function.\\$
$I= 2 \displaystyle \int_{{0}}^{\frac{\pi}{2}} sin^{2}\theta \: d\theta \\$
$= 2 \displaystyle \int_{{0}}^{\frac{\pi}{2}}\dfrac{1-cos2\theta}{2}\: d\theta \\$
$= \displaystyle \int_{{0}}^{\frac{\pi}{2}} 1-cos2\theta \: d\theta\\$
$= \bigg[\theta-\dfrac{sin2\theta}{2} \bigg]_{_{0}}^{\frac{\pi}{2}}\\$
$=\left (\dfrac{\pi}{2}-\dfrac{sin2\left ( \frac{\pi}{2} \right )}{2} \right )- \left (\dfrac{0-sin2(0)}{2} \right )\\$
$=\left (\dfrac{\pi}{2}-0 \right )- \left (0-0 \right )\\$
$=\dfrac{\pi}{2}\\$

$3. \displaystyle \int_{_{-1}}^{1} log\left (\dfrac{2-x}{2+x} \right )\: dx\\$

$I= \displaystyle \int_{_{-1}}^{1} log\left (\dfrac{2-x}{2+x} \right )\: dx\\$
$f(x)= log\left (\dfrac{2-x}{2+x} \right )\\$
$f(x)= log (2-x)-log (2+x)\\$
$f(-x)= log (2+x)-log (2-x)\\$
$=- \bigg[log (2-x)-log (2+x) \bigg] \\$
$f(-x)= - \bigg[log\left (\dfrac{2-x}{2+x} \right ) \bigg]\\$
$f(-x)=f(-x)\\$
$f(x) \: is \: an \: odd \: function.\\$
$\therefore I=0$

$4. \displaystyle \int_{_{0}}^{\frac{\pi}{2}} \frac{sin^{7}x}{sin^{7}x+cos^{7}x} \: dx\\$

$I= \displaystyle \int_{_{0}}^{\frac{\pi}{2}} \dfrac{sin^{7}x}{sin^{7}x+cos^{7}x} \: dx \cdots (1)\\$
$=\displaystyle \int_{_{0}}^{\frac{\pi}{2}} \dfrac{sin^{7}(\dfrac{\pi}{2}-x)}{sin^{7}(\dfrac{\pi}{2}-x)+cos^{7}(\dfrac{\pi}{2}-x)} \: dx \\$
$=\displaystyle \int_{_{0}}^{\frac{\pi}{2}} \dfrac{cos^{7}x}{cos^{7}x +sin^{7}x} \: dx \cdots (2)\\$
$=(1)+(2)\\$
$I+I=\displaystyle \int_{_{0}}^{\frac{\pi}{2}}\dfrac{sin^{7}x}{sin^{7}x+cos^{7}x}+\displaystyle \int_{_{0}}^{\frac{\pi}{2}}\dfrac{cos^{7}x}{cos^{7}x +sin^{7}x} \: dx\\$
$2I=\displaystyle \int_{_{0}}^{\frac{\pi}{2}}\dfrac{sin^{7}x+cos^{7}x}{sin^{7}x+cos^{7}x}\: dx\\$
$=\displaystyle \int_{_{0}}^{\frac{\pi}{2}} dx \\$
$=\bigg[x \bigg]_{_{0}}^{\frac{\pi}{2}}\\$
$2I=\dfrac{\pi}{2}-0\\$
$I=\dfrac{\pi}{4}$

$5. \displaystyle \int_{_{0}}^{1} log \left (\dfrac{1}{x}-1 \right )\: dx$

$I= \displaystyle \int_{_{0}}^{1} log \left (\dfrac{1}{x}-1 \right )\: dx\\$
$= \displaystyle \int_{_{0}}^{1} log \left (\dfrac{1-x}{x} \right )\: dx \cdots(1)\\$
$= \displaystyle \int_{_{0}}^{1} log \left (\dfrac{1-(1-x)}{1-x} \right )\: dx \\$
$= \displaystyle \int_{_{0}}^{1} log \left (\dfrac{1-1+x)}{1-x} \right )\: dx \\$
$= \displaystyle \int_{_{0}}^{1} log \left (\dfrac{x}{1-x} \right )\: dx\cdots (2) \\$
$(1)+(2)\\$
$I+I=\displaystyle \int_{_{0}}^{1} log \left (\dfrac{1-x}{x} \right )\: dx+\displaystyle \int_{_{0}}^{1} log \left (\dfrac{x}{1-x} \right )\: dx\\$
$2I=\displaystyle \int_{_{0}}^{1} \bigg[log \left (\dfrac{1-x}{x} \right )+ log \left (\dfrac{x}{1-x} \right ) \bigg]\: dx\\$
$=\displaystyle \int_{_{0}}^{1} log\left (\dfrac{1-x}{x} \right )\left (\dfrac{x}{1-x} \right )\:dx\\$
$=\displaystyle \int_{_{0}}^{1} log\: 1 \:dx\\$
$2I=0\\$
$I=0$

$6. \displaystyle \int_{_{0}}^{1} \dfrac{x}{(1-x)^{\frac{3}{4}}}\:dx\\$

$I=\displaystyle \int_{_{0}}^{1} \dfrac{x}{(1-x)^{\frac{3}{4}}}\:dx\\$
$=\displaystyle \int_{_{0}}^{1} \dfrac{(1-x)}{(1-(1-x))^{\frac{3}{4}}}\:dx\\$
$=\displaystyle \int_{_{0}}^{1} \dfrac{1-x}{x^{\frac{3}{4}}}\:dx\\$
$=\displaystyle \int_{_{0}}^{1} \dfrac{1}{x^{\frac{3}{4}}}-\frac{x}{x^{\frac{3}{4}}} \: dx\\$
$=\displaystyle \int_{_{0}}^{1} x^{\frac{-3}{4}}-x^{(1-\frac{3}{4})} \: dx\\$
$=\displaystyle \int_{_{0}}^{1} x^{\frac{-3}{4}}-x^{(\frac{1}{4})} \: dx\\$
$=\bigg[\dfrac{x^{\frac{-3}{4}+1}}{\frac{-3}{4}+1}-\frac{x^{(\frac{1}{4}+1)} }{\frac{1}{4}+1} \bigg]_{_{0}}^{1}\\$
$=\bigg[\dfrac{x^{\frac{1}{4}}}{\frac{1}{4}}-\frac{x^{\frac{5}{4}} }{\frac{5}{4}} \bigg]_{_{0}}^{1}\\$
$=\left [4x^{\frac{1}{4}}-\frac{4}{5}x^{\frac{5}{4}} \right ]_{_{0}}^{1}\\$
$=\bigg [4(1)-\frac{4}{5}(1) \bigg]- \bigg[0-0 \bigg]\\$
$=4-\frac{4}{5}\\$
$=\dfrac{16}{5}$