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
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