Integrals Test - 49

$$\displaystyle { I }_{ 1 }=\int _{ 1 }^{ 2 }{ \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } }  }  } dx$$

$$={ \left( \log { \left( x+\sqrt { 1+{ x }^{ 2 } }  \right)  }  \right)  }_{ 1 }^{ 2 }$$

$$\displaystyle =\log { \left( 2+\sqrt { 5 }  \right)  } -\log { \left( 1+\sqrt { 2 }  \right)  } =\log { \left( \frac { \left( 2+\sqrt { 5 }  \right)  }{ \left( 1+\sqrt { 2 }  \right)  }  \right)  } $$

$$\displaystyle { I }_{ 2 }=\int _{ 1 }^{ 2 }{ \frac { 1 }{ x }  } dx={ \left( \log { x }  \right)  }_{ 1 }^{ 2 }=\log { 2 } -\log { 1 } =\log { 2 } $$
$$\therefore { I }_{ 2 }>{ I }_{ 1 }$$

Let, $$z=e^{x^2}\implies dz=2xe^{x^2}dx$$

Let us take $$\displaystyle \frac{x^{m}}{x^{(p+q)}+1}=t$$ where m is either p or q as seen from the options,

$$\displaystyle \frac{\mathrm{d} t}{\mathrm{d} x}=\frac{(x^{(p+q)}+1)mx^{m-1}-x^{m}(p+q)x^{p+q-1}}{x^{2p+2q}+1+2x^{p+q}}$$
$$\displaystyle =\frac{(mx^{(p+q+m-1)}+mx^{m-1}-px^{p+q+m-1}-qx^{p+q+m-1})}{x^{2p+2q}+1+2x^{p+q}}$$
we want powers of $$p+2q-1$$ and $$q-1$$, so $$m=q,$$
$$\displaystyle =\frac{(qx^{(p+2q-1)}+qx^{q-1}-px^{p+2q-1}-qx^{p+2q-1})}{x^{2p+2q}+1+2x^{p+q}}$$
$$\displaystyle =\frac{qx^{q-1}-px^{p+2q-1}}{x^{2p+2q}+1+2x^{p+q}}$$
$$\displaystyle =\frac{\mathrm{d} t}{\mathrm{d} x}$$
so the integral will be $$-\dfrac{x^{q}}{x^{(p+q)}+1}$$

Let $$\displaystyle I=\int _{ 1 }^{ e^{ 37 } } \dfrac { \pi \sin  \left( \pi \log _{ e } x \right)  }{ x } dx$$
Substitute $$\displaystyle \pi \log { x } =t\Rightarrow \dfrac { \pi  }{ x } dx=dt$$

Let $$\displaystyle I=\sqrt { \frac { 2+x }{ 2-x }  } dt$$
Put $$\displaystyle t=\frac { 2+x }{ 2-x } \Rightarrow dt=\left( \frac { 1 }{ 2-x } -\frac { -x-2 }{ { \left( 2-x \right)  }^{ 2 } }  \right) dx$$
$$\displaystyle \therefore I=4\int { \frac { \sqrt { t }  }{ { \left( t+1 \right)  }^{ 2 } } dt } $$

Put $$\displaystyle u=\sqrt { t } \Rightarrow du=\frac { 1 }{ 2\sqrt { t }  } dt$$
$$\displaystyle \therefore I=8\int { \frac { { u }^{ 2 } }{ { \left( { u }^{ 2 }+1 \right)  }^{ 2 } }  } du=8\int { \left( \frac { 2 }{ { u }^{ 2 }+1 } -\frac { 1 }{ { \left( { u }^{ 2 }+1 \right)  }^{ 2 } }  \right) du } $$

$$\displaystyle =\frac { 4\left( \left( { u }^{ 2 }+1 \right) \tan ^{ -1 }{ u-u }  \right)  }{ { u }^{ 2 }+1 } +C$$
$$\displaystyle =\frac { 4\left( \left( t+1 \right) \tan ^{ -1 }{ \left( \sqrt { t }  \right) -\sqrt { t }  }  \right)  }{ t+1 } +C$$
$$\displaystyle =\sqrt { \frac { 2+x }{ 2-x }  } \left( x-2 \right) +4\tan ^{ -1 }{ \left( \sqrt { \frac { 2+x }{ 2-x }  }  \right)  } +C$$

Therefore
$$\displaystyle \int _{ 0 }^{ 2 }{ \sqrt { \frac { 2+x }{ 2-x }  } dx } =\left[ \sqrt { \frac { 2+x }{ 2-x }  } \left( x-2 \right) +4\tan ^{ -1 }{ \left( \sqrt { \frac { 2+x }{ 2-x }  }  \right)  }  \right] _{ 0 }^{ 2 }$$
$$\displaystyle =\left[ -2+4\frac { \pi  }{ 2 } - \right] -0=2\pi -2$$

$$I=\int _{ 0 }^{ \infty  }{ f\left( x+\dfrac { 1 }{ x }  \right) .\dfrac { \ln { x }  }{ x } dx } $$
Let $$x={ e }^{ t }$$
$$\Rightarrow dx={ e }^{ t }dt$$
$$I=\int _{ -\infty  }^{ \infty  }{ f\left( { e }^{ t }+{ e }^{ -t } \right) tdt } =0$$        since, $$f\left( { e }^{ t }+{ e }^{ -t } \right) t$$ is an odd function.

Ans: 0

Let $$\displaystyle I=\int_0^{\displaystyle\frac{\pi}{4}}{\frac{\sin{x}+\cos{x}}{9+16\{1-{(\sin{x}-\cos{x})}^2\}}dx}$$
$$\displaystyle\therefore I=\int_0^{\displaystyle\frac{\pi}{4}}{\frac{\sin{x}+\cos{x}}{25-16{(\sin{x}-\cos{x})}^2}dx}$$,
Substitute $$4(\sin{x}-\cos{x})=t\implies4(\cos{x}+\sin{x})dx=dt$$
$$\displaystyle\frac{1}{4}\int_{\displaystyle-4}^0{\frac{dt}{25-t^2}}$$
$$\displaystyle I=\frac{1}{4}.\frac{1}{2(5)}{\left(\log{\left|\frac{5-t}{5+t}\right|}\right)}_{\displaystyle-4}^0$$
$$\displaystyle=\frac{1}{40}\left[\log{\left|\frac{5-0}{5+0}\right|}-\log{\left|\frac{5+4}{5-4}\right|}\right]$$
$$\displaystyle=\frac{1}{40}\left[\log{1}-\log{\left(\frac{1}{9}\right)}\right]$$, (where $$\log{1}=0$$)
$$\displaystyle=\frac{1}{40}[\log{9}]=\frac{2}{40}\log{3}=\frac{1}{20}(\log{3})$$

Since, $$\displaystyle 2f(x)+f(-x)=\frac{1}{x}\sin{\left(x-\frac{1}{x}\right)}$$   ...(i)
$$\displaystyle\therefore 2f(-x)+f(x)=\frac{1}{x}\sin{\left(x-\frac{1}{x}\right)}$$ (replace $$x$$ by $$-x$$)   ...(ii)
$$\implies f(x)=f(-x)$$ [subtracting equations (i) and (ii)]
$$\displaystyle\therefore 3f(x)=\frac{1}{x}\sin{\left(x-\frac{1}{x}\right)}$$
Hence,

$$\displaystyle I=\int_{\frac{1}{e}}^{e}{f(x)dx}=\frac{1}{3}\int_{\frac{1}{e}}^{e}{\frac{1}{x}\sin{\left(x-\frac{1}{x}\right)}dx}$$
Now, put $$\displaystyle x=\frac{1}{t}\Rightarrow\displaystyle dx=-\frac{1}{t^2}dt$$
$$\displaystyle\therefore I=\frac{1}{3}\int_{e}^{\frac{1}{e}}{t\sin{\left(\frac{1}{t}-t\right)}.\left(-\frac{1}{t^2}\right)dt}$$
$$\displaystyle=\frac{1}{3}\int_{e}^{\frac{1}{e}}{\frac{1}{t}\sin{\left(t-\frac{1}{t}\right)}dt}$$
$$\displaystyle=-\frac{1}{3}\int_{\frac{1}{e}}^{e}{\frac{1}{t}\sin{\left(t-\frac{1}{t}\right)}dt}$$
$$\Rightarrow I=-I\Rightarrow 2I=0\Rightarrow I=0$$
$$\displaystyle\therefore\int_{\frac{1}{e}}^{e}{f(x)dx}=0$$.

$$\displaystyle \left( \int _{ 0 }^{ 1 }{ \frac { dt }{ t^{ 2 }+2t\cos { \alpha  } +1 }  }  \right) x^{ 2 }-\left( \int _{ -3 }^{ 3 }{ \frac { t^{ 2 }\sin { 2t }  }{ t^{ 2 }+1 } dt }  \right) x-2=0$$

Here, $$\displaystyle \int _{ -3 }^{ 3 }{ \frac { t^{ 2 }\sin { 2t }  }{ t^{ 2 }+1 } dt } =0 \left(\because f(t)=\frac { t^{ 2 }\sin { 2t }  }{ t^{ 2 }+1 } \text{ is odd}\right)$$

So, the equation becomes
$$\displaystyle \left( \int _{ 0 }^{ 1 }{ \frac { dt }{ t^{ 2 }+2t\cos { \alpha  } +1 }  }  \right) x^{ 2 }-2=0$$   ....(1)

Now,

Given , at $$x=a , \frac{dy}{dx}=\tan {\frac{\pi}{3}}=\sqrt{3}$$
$$\Rightarrow f'(a)=\sqrt{3}$$
Also, at $$x=b, \frac{dy}{dx}=\tan {\frac{\pi}{4}}=1$$
$$\Rightarrow f'(b)=1$$
Now, $$\displaystyle \int_{a}^{b} f'(x).f''(x)dx$$ 
Put $$f'(x)=t \Rightarrow f''(x)dx=dt$$
        $$=\displaystyle \int_{\sqrt{3}}^{1} t dt$$
        $$=-\displaystyle \int_{1}^{\sqrt{3}} t dt$$

        $$\displaystyle =-[\frac{t^{2}}{2}]_{1}^{\sqrt{3}}$$

        $$=-1$$