Mathematics

# a) Prove that $\displaystyle \overset {a}{ \underset {a}{\int}}f(x)dx=\begin{cases}2\overset { a }{ \underset { 0 }{\displaystyle \int } } f(x)dx & \text{if }f (x) \text{ is an even function} \\ 0 & \text{if }f (x)\text{ is an odd function} \end{cases}$and hence evaluate $\displaystyle\overset { 1 }{ \underset { -1 }{\int} } {\sin}^5 x {\cos}^4 x dx$.b) Prove that$\begin{vmatrix}a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1\end{vmatrix} = 1+a^2+b^2+c^2$.

##### SOLUTION
$(a)$
$\displaystyle \int_{-a}^{a} f(x) dx=\int_{-a}^{0} f(x) dx + \int_{0}^{a} f(x) dx$
Put $x=-t$ in first integral $\implies dx=-dt$
When $x=-a, t=a$
$x=0, t=0$
$\displaystyle \int_{-a}^{a} f(x) dx=-\int_{a}^{0} f(-t) dt + \int_{0}^{a} f(x) dx$
$=\displaystyle \int_{0}^{a} f(t) dt + \int_{0}^{a} f(x) dx$ ....... (Since, $f$ is even)
$=\displaystyle \int_{0}^{a} f(x) dx + \int_{0}^{a} f(x) dx$
$=\displaystyle 2\int_{0}^{a} f(x) dx$

$\displaystyle \int_{-a}^{a} f(x) dx=-\int_{a}^{0} f(-t) dt + \int_{0}^{a} f(x) dx$
$=-\displaystyle \int_{0}^{a} f(t) dt + \int_{0}^{a} f(x) dx$ ..... (Since, $f$ is odd. So, $f(-t)=-f(t)$)
$=-\displaystyle \int_{0}^{a} f(x) dx + \int_{0}^{a} f(x) dx$
$=0$
To evaluate $\displaystyle \int_{-1}^{1} \sin^5 x \cos^4 x dx$
$\sin(-x)=-\sin x$ and $\cos(-x)=\cos x$
$\implies \sin x$ is odd and $\cos x$ is even
$\implies \sin^5 x$ is odd and $\cos^4 x$ is even
$\implies f(x)=\sin^5 x \cos^4 x$ is odd
$\implies \displaystyle \int_{-1}^{1} \sin^5 x \cos^4 x dx=0$ ...... ($\because f(x)=\sin^5 x \cos^4 x$ is an odd function)

$(b)$
Consider, $\begin{vmatrix} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1 \end{vmatrix}$

$=(a^2+1)\begin{vmatrix} b^2+1 & bc \\ cb & c^2+1\end{vmatrix}-ab\begin{vmatrix} ab & bc \\ ca & c^2+1\end{vmatrix}+ac\begin{vmatrix} ab & b^2+1 \\ ca & cb \end{vmatrix}$

$=(a^2+1)[b^2c^2+b^2+c^2+1-cbbc] - ab[abc^2+ab-abc^2] + ac[ab^2c-ab^2c-ac]$

$=(a^2+1)[b^2+c^2+1] - ab[ab] + ac[-ac]$
$=a^2b^2+a^2c^2+a^2+b^2+c^2+1-a^2b^2-a^2c^2$
$=a^2+b^2+c^2+1$

Hence, $\begin{vmatrix} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1 \end{vmatrix}=1 + a^2 + b^2 + c^2$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

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$\displaystyle \int_{0}^{\pi /2} \displaystyle \frac{1}{\sin x+\cos x} \ dx$
• A. $\sqrt{2}l\mathrm{o}\mathrm{g}(\sqrt{2}-1)$
• B. $\dfrac{1}{\sqrt{2}}l\mathrm{o}\mathrm{g}(\sqrt{2}+1)$
• C. $\dfrac{-1}{\sqrt{2}}l\mathrm{o}\mathrm{g}(\sqrt{2}+1)$
• D. $\sqrt{2}l\mathrm{o}\mathrm{g}(\sqrt{2}+1)$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
The value of $\displaystyle \int_{-4}^{-5}e^{\left ( x+5 \right )^{2}}dx+3\displaystyle \int_{1/3}^{2/3}e^{9\left ( x-2/3 \right )^{2}}dx$ is
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• C. $1/2$
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1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
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Q4 Single Correct Medium
$\displaystyle \int e^{x}\left ( \frac{2\tan x}{1+x}+\tan ^{2}\left ( x+\frac{\pi }{4} \right ) \right )dx$ is equal to
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Evaluate $\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}}dx$