The graph of y = sin x is shown in the figure below
Clearly, required region is bounded by y = sin x, x-axis and lines x = 0 and x = 2\(\pi\), which is represented by shaded region.
\(\therefore\) Required area = Area of region OABO + Area of region BCDB
\(=\int_{0}^{\pi} \sin x d x+\mid \int_{\pi}^{2 \pi} \sin x d x\) 
[since, in region BCDB, the graph is below the X-axis, so its area comes out to be negative, therefore we take the absolute value]
\(=[-\cos x]_{0}^{\pi}+\left|[-\cos x]_{\pi}^{2 \pi}\right|\) 
\(=(-\cos \pi+\cos 0)+|-\cos 2 \pi+\cos \pi|\) 
\(=(1+1)+|-1-1|\) 
\(=2+|-2|=2+2=4 \text { sq units }\) 
Hence, the required area is 4 sq. units.

\(I=\quad \int { \frac { 1 }{ { x }^{ 4 }-1 } dx. } \)
Now  \(\frac { 1 }{ { x }^{ 4 }-1 } =\frac { 1 }{ (x-1)(x+1)({ x }^{ 2 }+1) } .\)
Let  \(\frac { 1 }{ { x }^{ 4 }-1 } \equiv \frac { A }{ x-1 } +\frac { B }{ x+1 } +\frac { Cx+D }{ { x }^{ 2 }+1 } \quad \quad \quad .......(1)\)
Multiplying by \({ x }^{ 4 }-1\)  
we get: \(1\equiv A(x+1)({ x }^{ 2 }+1)+B(x-1)({ x }^{ 2 }+1)+C(Cx+D)(x-1)(x+1).\)
Putting \(x=1,1=A(2)(2)\Rightarrow A=-\frac { 1 }{ 4 } .\)
Putting \(x=-1,1=B(-2)(2)\Rightarrow B=-\frac { 1 }{ 4 } .\)
Putting  \(x=0,1=A-B-D\)
\(\Rightarrow D=A-B-1=\frac { 1 }{ 4 } +\frac { 1 }{ 4 } -1=-\frac { 1 }{ 2 } .\)
Comparing coeffs.of  \({ x }^{ 3 },0, 1=A+B+C\)
\(\Rightarrow C=-A-B=-\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =0\)
Putting in (1),
\(\frac { 1 }{ { x }^{ 4 }-1 } =\frac { \frac { 1 }{ 4 } }{ x-1 } +\frac { -\frac { 1 }{ 4 } }{ x+1 } +\frac { -\frac { 1 }{ 2 } }{ { x }^{ 2 }+1 } \)
\(\therefore I= \int { \frac { 1 }{ { x }^{ 4 }-1 } dx. } \)
\(=\frac { 1 }{ 4 } \int { \frac { 1 }{ x+1 } } dx-\frac { 1 }{ 4 } \int { \frac { 1 }{ x+1 } } dx-\frac { 1 }{ 2 } \int { \frac { dx }{ { 1 }^{ 2 }+{ x }^{ 2 } } } \)
\(=\frac { 1 }{ 4 } \log { \left| x-1 \right| } -\frac { 1 }{ 4 } \log { \left| x+1 \right| } -\frac { 1 }{ 2 } \tan ^{ -1 }{ \frac { x }{ 1 } } +C\)
\(=\frac { 1 }{ 4 } \log { \left| \frac { x-1 }{ x+1 } \right| } -\frac { 1 }{ 2 } \tan ^{ -1 }{ x } +C.\)

\(\int \frac{x-1}{\sqrt{x^{2}-1}} d x=\int \frac{x}{\sqrt{x^{2}-1}} d x-\int \frac{1}{\sqrt{x^{2}-1}} d x \)
\(\text { For } \int \frac{x}{\sqrt{x^{2}-1}} d x, \text { let } x^{2}-1=t \Rightarrow 2 x d x=d t \)
\(\therefore \int \frac{x}{\sqrt{x^{2}-1}} d x =\frac{1}{2} \int \frac{d t}{\sqrt{t}} \)
\(=\frac{1}{2} \int t^{-\frac{1}{2}} d t \)
\(=\frac{1}{2}\left[2 t^{\frac{1}{2}}\right] \)
\(=\sqrt{t} \)
\(=\sqrt{x^{2}-1}\)
From (1), we obtain
\(\int \frac{x-1}{\sqrt{x^{2}-1}} d x =\int \frac{x}{\sqrt{x^{2}-1}} d x-\int \frac{1}{\sqrt{x^{2}-1}} d x \quad\left[\int \frac{1}{\sqrt{x^{2}-a^{2}}} d t=\log \left|x+\sqrt{x^{2}-a^{2}}\right|\right] \)
\(=\sqrt{x^{2}-1}-\log \left|x+\sqrt{x^{2}-1}\right|+\mathrm{C} \)

\(I=\int { \frac { { e }^{ 2x }-1 }{ { e }^{ 2x }+1 } } dx\)
\(=\int { \frac { { e }^{ x }\left( { e }^{ x }-{ e }^{ -x } \right) }{ { e }^{ x }\left( { e }^{ x }+{ e }^{ -x } \right) } } dx\quad =\int { \frac { \left( { e }^{ x }-{ e }^{ -x } \right) }{ \left( { e }^{ x }+{ e }^{ -x } \right) } dx } \)
Put  \({ e }^{ x }+{ e }^{ -x }=t\) so that \(({ e }^{ x }-{ e }^{ -x })\ dx=dt.\)
\(\therefore I=\int { \frac { dt }{ t } } =\log { \left| t \right| } +C\)
\(=\log { \left| { e }^{ x }+{ e }^{ -x } \right| } +C\) 

Given, differential equation is \(\begin{aligned} \frac{d y}{d x} & =1+x^2+y^2+x^2 y^2 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{d y}{d x} & =1\left(1+x^2\right)+y^2\left(1+x^2\right) \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{d y}{d x} & =\left(1+x^2\right)\left(1+y^2\right) \end{aligned}\)
On separating the variables, we get \(\Rightarrow \frac{d y}{1+y^2}=\left(1+x^2\right) d x\)
On integrating both sides, we get \(\int \frac{d y}{1+y^2}=\int\left(1+x^2\right) d x\)
\(\Rightarrow \quad \tan ^{-1} y=x+\frac{x^3}{3}+C\)          ...(i)
Also, given that y = 1, when x = 0.
On putting x = 0 and y = 1 in Eq. (i), we get
tan^-1 1 = C
\(\Rightarrow \quad \tan ^{-1}(\tan \pi / 4)=C \quad\left[\because \tan \frac{\pi}{4}=1\right]\)
\(\Rightarrow \quad C=\pi / 4\)
On putting the value of C in Eq. (i), we get
\(\begin{aligned} \tan ^{-1} y & =x+\frac{x^3}{3}+\frac{\pi}{4} \end{aligned}\)
\(\begin{aligned} \therefore \quad y & =\tan \left(x+\frac{x^3}{3}+\frac{\pi}{4}\right) \end{aligned}\)
which is the required solution.

Let radius of cone be x and its height be h.
\(\therefore\) OD = (h - r)

Volume of cone (V)
\(=\frac { 1 }{ 3 } \pi { x }^{ 2 }h\) ...(i)
In \(\Delta OCD,\quad { x }^{ 2 }+({ h-r) }^{ 2 }={ r }^{ 2 }or\quad { x }^{ 2 }={ r }^{ 2 }-{ (h-r) }^{ 2 }\)
\(\therefore V=\frac { 1 }{ 3 } \pi h\{ { r }^{ 2 }-(h-r{ ) }^{ 2 }\} \)
\(=\frac { 1 }{ 3 } \pi (-{ h }^{ 3 }+{ 2h }^{ 2 }r)\)
\(\Rightarrow \frac { dV }{ dh } =\frac { \pi }{ 3 } (-3{ h }^{ 2 }+4hr)\)
\(\therefore \quad \frac { dV }{ dh } =0\Rightarrow h=\frac { 4r }{ 3 } \)
\(\frac { { d }^{ 2 }V }{ { dh }^{ 2 } } =\frac { \pi }{ 3 } (-6h+4r)\)
\(=\frac { \pi }{ 3 } \left( -6\left( \frac { 4r }{ 3 } \right) +4r \right) \)
\(=-\frac { 4\pi r }{ 3 } <0\)
\(\therefore \ at\quad h=\frac { 4r }{ 3 } \), Volume is maximum
Maximum volume
\(=\frac { 1 }{ 3 } \pi .\left\{ -{ \left( \frac { 4r }{ 3 } \right) }^{ 3 }+2{ \left( \frac { 4r }{ 3 } \right) }^{ 2 }r \right\} \)
\(=\frac { 8 }{ 27 } .\left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) \)
\(=\frac { 8 }{ 27 } \) (volume of sphere)

\(I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { dx }{ 1+\sqrt { \cot { x } } } } \)
\(=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \sin { x } } }{ \sqrt { \sin { x } } +\sqrt { \cos { x } } } } dx\)
Apply property \(\int _{ a }^{ b }{ f\left( x \right) } =\int _{ a }^{ b }{ f\left( a+b-x \right) } \)
\(=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \sin { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } }{ \sqrt { \sin { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } +\sqrt { \cos { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } } } dx\)
\(\therefore I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \cos { x } } }{ \sqrt { \cos { x } } +\sqrt { \sin { x } } } } dx\)
Adding, we get \(2I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ dx } =\left[ x \right] _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }\)

Area (EABCDE) = Area (ABFE) + Area (CBFD)
​​​\(=\int _{ -3 }^{ -1 }{ \left( \left| x+1 \right| +1 \right) dx+\int _{ -1 }^{ 3 }{ \left( \left| x+1 \right| +1 \right) dx } } \)
\(=\int _{ -3 }^{ -1 }{ \left( -x \right) dx } +\int _{ -1 }^{ 3 }{ \left( x+2 \right) dx } \)
\(=\left[ -\frac { { x }^{ 2 } }{ 2 } \right] _{ -3 }^{ -1 }+\left[ \frac { { \left( x+2 \right) }^{ 2 } }{ 2 } \right] _{ -1 }^{ 3 }\)
\(=-\frac { 1 }{ 2 } \left( 1-9 \right) +\frac { 1 }{ 2 } \left( 25-1 \right) =16\ sq.unit\)

We have, the equation of the ellipse
x² + 9y² = 36
\(\Rightarrow \frac{x^2}{36}+\frac{y^2}{4}=1 \Rightarrow \frac{x^2}{6^2}+\frac{y^2}{2^2}=1\)

\(\therefore\) Required area \(=4 \int_0^6 y d x\)
\(\begin{aligned} & =4 \int_0^6 \frac{2}{6} \sqrt{36-x^2} d x=\frac{4}{3} \int_0^6 \sqrt{36-x^2} d x \end{aligned}\)
\(\begin{aligned} =\frac{4}{3}\left[\frac{x}{2} \sqrt{36-x^2}+\frac{36}{2} \sin ^{-1} \frac{x}{6}\right]_0^6 \end{aligned}\)
\(\begin{aligned} & =\frac{4}{3}\left[0+\frac{36}{2} \sin ^{-1} 1-0\right]=\frac{4}{3} \times 18 \times \frac{\pi}{2} \end{aligned}\)
= 12 \(\pi\) sq units

\(y=\sqrt{a^{2}-x^{2}}\)
Differentiating both sides of this equation with respect to x, we get:
\(\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{a^{2}-x^{2}}\right) \)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}} \cdot \frac{d}{d x}\left(a^{2}-x^{2}\right) \)
\(=\frac{1}{2 \sqrt{a^{2}-x^{2}}}(-2 x) \)
\(=\frac{-x}{\sqrt{a^{2}-x^{2}}}\)
Substituting the value of \( \frac{d y}{d x}\) in the given differential equation, we get:
\(\text { L.H.S. } =x+y \frac{d y}{d x} =x+\sqrt{a^{2}-x^{2}} \times \frac{-x}{\sqrt{a^{2}-x^{2}}} \)
\(=x-x =0 =\text { R.H.S }\)
Hence, the given function is the solution of the corresponding differential equation. \(\)

Let \(I=\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x=\int \frac{e^{\log x^{\circ}}-e^{\log x^{3}}}{e^{\log x^{4}}-e^{\log x^{3}}} d x\) 
\(=\int \frac{x^{6}-x^{5}}{x^{4}-x^{3}} d x \quad\left[\because e^{\log f(x)}=f(x)\right][1]\)
\(=\int \frac{x^{5}(x-1)}{x^{3}(x-1)} d x=\int x^{2} d x=\frac{x^{3}}{3}+C \) 

\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log\ tanxdx } \)..(i)
Apply the property  \(\int _{ 0 }^{ a }{ f(x) } =\int _{ 0 }^{ a }{ f(a-x)dx, } \)
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log\quad tan\left( \frac { \pi }{ 2 } -x \right) dx } \)  ...(ii)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log\quad cotxdx } \)
by adding eqn. (i) and (ii)
\(2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ (log\quad tanx+log\quad cotx)xdx } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log(tanx\times cotx } )dx\)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log1dx } =0\)
\(\Rightarrow 2I=0\)
\(\Rightarrow I=0\)