\(= {sin^4\ x\over4}-{sin^6\ x\over 6}+c\)

\(tan\ x-cot\ x-3x+c\)

The given parabola is x² = y
This is an upward parabola with vertex (0, 0)
y = IxI represents the st. lines:
y = x and y = -x
y = x meets (1) at O (0, 0) and A (1, 1)
y = -x meets (1) at O (0, 0) and A (-1, 1)
Therefore, Required area = 2 (Shaded area in first quadrant)
\(=2\left( \overset { 1 }{ \underset { 0 }{ \int { } } } xdx-\overset { 1 }{ \underset { 0 }{ \int { } } } { x }^{ 2 }dx \right) \)
\(=2\left( \left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 1 }-\left[ \frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 } \right) \)
\(=2\left( \left( \frac { 1 }{ 2 } -0 \right) -\left( \frac { 1 }{ 3 } -0 \right) \right) =2\left( \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right) \)
\(=2\left( \frac { 1 }{ 6 } \right) =\frac { 1 }{ 3 } sq.unit.\)

Let  \(\int _{ 0 }^{ \pi /4 }{ (\sqrt { \tan { x } } +\sqrt { \cot { x } } ) } dx\)
\(=\int _{ 0 }^{ \pi /4 }{ \frac { \tan { x+1 } }{ \sqrt { \tan { x } } } } dx\)
\(Put \sqrt { \tan { x } } =t\ i.e. \tan { x } x={ t }^{ 2 }\)
so that \(\sec ^{ 2 }{ x } dx=2t\ dt \Rightarrow dx=\frac { 2t }{ 1+{ t }^{ 4 } } dt.\)
When \(x=0, t=0.\) When \(x=\frac { \pi }{ 4 } ,t=1.\)
\(\therefore I= \int _{ 0 }^{ 1 }{ \frac { { t }^{ 2 }+1 }{ t } } .\frac { 2t }{ 1+{ t }^{ 4 } } dt\)
\(=2\int _{ 0 }^{ 1 }{ \frac { { t }^{ 2 }+1 }{ { t }^{ 4 }+1 } } dx=2\int _{ 0 }^{ 1 }{ \frac { 1+1/{ t }^{ 2 } }{ { t }^{ 2 }+1/{ t }^{ 2 } } } dt\)
\(Put\ t-\frac { 1 }{ t } =y\) so that \(\left( 1+\frac { 1 }{ { t }^{ 2 } } \right) dt=dy.\)
Also \({ t }^{ 2 }-2+\frac { 1 }{ { t }^{ 2 } } ={ y }^{ 2 } \Rightarrow { t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } ={ y }^{ 2 }+2\)
\(\therefore I=2\int _{ t=0 }^{ 1 }{ \frac { dy }{ { y }^{ 2 }+2 } } \)
\(=\frac { 2 }{ \sqrt { 2 } } { \left[ \tan ^{ -1 }{ \frac { y }{ \sqrt { 2 } } } \right] }_{ t=0 }^{ 1 }\)
\(=\sqrt { 2 } { \left[ \tan ^{ -1 }{ \frac { 1 }{ \sqrt { 2 } } } \left( t-\frac { 1 }{ t } \right) \right] }_{ 0 }^{ 1 }\)
\(=\sqrt { 2 } \left[ \tan ^{ -1 }{ (0) } -\tan ^{ -1 }{ (-\infty ) } \right] \)
\(=\sqrt { 2 } \left[ \tan ^{ -1 }{ (-\infty ) } \right] =\sqrt { 2 } .\frac { \pi }{ 2 } .\)

We have two circles
\(x^{2}+y^{2}=4\)   ... (i)
whose centre is origin and radius is 2;
and \((x-2)^{2}+y^{2}=4\) 
whose centre is (2, 0) and radius is 2
 
Here, region enclosed between two circles is OACDO,
which is symmetrical about X-axis
Now, from (i) and (ii), we get
\( (x-2)^{2}+y^{2}=x^{2}+y^{2} \)
\(\Rightarrow x^{2}+4-4 x+y^{2}=x^{2}+y^{2} \)
\(\Rightarrow 4-4 x=0 \Rightarrow x=1 \) 
From (i), we get
\(y^{2}=4-x^{2}=4-1=3 \Rightarrow y=\pm \sqrt{3}\) 
So, the points of intersection are \(A(1, \sqrt{3}) \text { and } D(1,-\sqrt{3})\) .
Here, we will draw two vertical strips [one for circle (i) and other for circle (ii)].
\(\therefore\) Required area = Area of region OACDO
= 2 x Area of region OACO
= 2 [Area of region OAMO + Area of region AMCA]
\(=2\left[\int_{0}^{1} y(\text { circle }(\mathrm{ii})) d x+\int_{1}^{2} y(\text { circle }(\mathrm{i})) d x\right]\) 
\(=2\left[\int_{0}^{1} \sqrt{4-(x-2)^{2}} d x+\int_{1}^{2} \sqrt{4-x^{2}} d x\right]\) 
\(\left[\begin{array}{l} \text { from Eq. }(\mathrm{i}), y^{2}=4-x^{2} \Rightarrow y=\sqrt{4-x^{2}} \\ \text { and from Eq. (ii), } y^{2}=4-(x-2)^{2} \\ \Rightarrow y=\sqrt{4-(x-2)^{2}} \end{array}\right]\) 
\( =2\left[\frac{(x-2)}{2} \sqrt{4-(x-2)^{2}}+\frac{4}{2} \sin ^{-1}\left(\frac{x-2}{2}\right)\right]_{0}^{1} +2\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{1}^{2} \) 
\( =\left[(x-2) \sqrt{4-(x-2)^{2}}\right.\left.+4 \sin ^{-1}\left(\frac{x-2}{2}\right)\right]_{0}^{1} +\left[x \sqrt{4-x^{2}}+4 \sin ^{-1} \frac{x}{2}\right]_{1}^{2} \) 
\( =\left[\left\{-\sqrt{3}+4 \sin ^{-1}\left(\frac{-1}{2}\right)\right\}-0-4 \sin ^{-1}(-1)\right] +\left[0+4 \sin ^{-1} 1-\sqrt{3}-4 \sin ^{-1} \frac{1}{2}\right] \) 
\(=\left[\left\{-\sqrt{3}-4 \times \frac{\pi}{6}\right\}+4 \times \frac{\pi}{2}\right]+\left[4 \times \frac{\pi}{2}-\sqrt{3}-4 \times \frac{\pi}{6}\right]\) 
\(=\left(-\sqrt{3}-\frac{2 \pi}{3}+2 \pi\right)+\left(2 \pi-\sqrt{3}-\frac{2 \pi}{3}\right)\) 
\(=\left(\frac{8 \pi}{3}-2 \sqrt{3}\right) \text { sq units }\) 
Hence, the required area is \(\left(\frac{8 \pi}{3}-2 \sqrt{3}\right) \text { sq units }\)

Let \(I=\int \frac{d x}{\sqrt{3-2 x-x^2}}\)
\(\therefore  I  =\int \frac{d x}{\sqrt{3-\left(2 x+x^2\right)}} \)
\(=\int \frac{d x}{\sqrt{3-\left(1+2 x+x^2-1\right)}}\)
\(=\int \frac{d x}{\sqrt{3-\left[(x+1)^2-1\right]}}\)
\(=\int \frac{d x}{\sqrt{3-(x+1)^2+1}}\)
\(\Rightarrow \quad I=\int \frac{d x}{\sqrt{4-(x+1)^2}}\)
\(\text { Let }(x+1)=t \Rightarrow d x=d t\)
\(\therefore \quad I=\int \frac{d t}{\sqrt{4-t^2}}=\int \frac{d t}{\sqrt{(2)^2-t^2}}\)
\(\Rightarrow I=\sin ^{-1}\left(\frac{t}{2}\right)+C\)
\( {\left[\because \int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1}\left(\frac{x}{a}\right)+C\right]} \)
\(\therefore \quad I=\sin ^{-1}\left(\frac{x+1}{2}\right)+C\) \([\because t=x+1]\)

\(\int { sin^{ 2 }x } dx=\int { \left( \frac { 1-cos2x }{ 2 } \right) dx } \)
\([\therefore cos2x=1-2sin^{ 2 }x]\)
\(=\frac { 1 }{ 2 } \int { dx } -\frac { 1 }{ 2 } \int { cos2x } dx\)
\(=\frac { 1 }{ 2 } x-\frac { 1 }{ 4 } sin2x+C\)

Let  \(I =\int\left(3 \operatorname{cosec}^{2} x-5 x+\sin x\right) d x \)  
\(=3 \int \operatorname{cosec}^{2} x d x-5 \int x d x+\int \sin x d x \)
\(=3(-\cot x)-\frac{5 x^{2}}{2}-\cos x+C \)
\(=-3 \cot x-\frac{5 x^{2}}{2}-\cos x+C\)

\(\frac{16}{3} \text { sq units }\)

\(\text { Required area }=\int_{0}^{2} y^{3} d x+\left|\int_{-1}^{0} y^{3} d x\right| \)

\(= \frac{17}{4} \text { sq units } \)

Anti-derivative of \(\left(3 \sqrt{x}+\frac{1}{\sqrt{x}}\right)\)
\(=\int\left(3 \sqrt{x}+\frac{1}{\sqrt{x}}\right) d x=3 \int \sqrt{x} d x+\int \frac{1}{\sqrt{x}} d x\)
\(=3\left(\frac{x^{(1 / 2)+1}}{(1 / 2)+1}\right)+\left[\frac{x^{(-1 / 2)+1}}{(-1 / 2)+1}\right]+C\)
\(=2\left(x^{3 / 2}+x^{1 / 2}\right)+C\)

\(I=\int { \frac { { x }^{ 2 }+x+1 }{ (x+2)({ x }^{ 2 }+1) } } dx\)
\(=\int { \left[ \frac { A }{ (x+2) } +\frac { Bx+C }{ ({ x }^{ 2 }+1) } \right] dx } \)
\(\Rightarrow A{ x }^{ 2 }+A+B{ x }^{ 2 }+2Bx+Cx+2C\)
\(=x+{ x }^{ 2 }+1\)
\(A+B=1\)
\(2B+C=1\)
\(A+2C=1\)
\(A=3/5,\)
\(B=2/5,\)
and ​​​​​​\(C=1/5\)
\(\therefore I=\int { \left[ \frac { 3 }{ 5(x+2) } +\frac { \frac { 2 }{ 5 } x+\frac { 1 }{ 5 } }{ ({ x }^{ 2 }+1 } \right] dx } \)
\(=\frac { 1 }{ 5 } \int { \left[ \frac { 3 }{ (x+2) } +\frac { 2x+1 }{ ({ x }^{ 2 }+1) } \right] } dx\)
\(=\frac { 1 }{ 5 } \left[ \int { \frac { 3 }{ (x+2) } } dx+\int { \frac { 2x+1 }{ ({ x }^{ 2 }+1) } dx+\int { \frac { 1 }{ ({ x }^{ 2 }+1) } dx } } \right] \)
\({ I }_{ 1 }=\frac { 1 }{ 5 } \int { \frac { 3 }{ (x+2) } } dx\)
\(=\frac { 3 }{ 5 } log(x+2)+{ C }_{ 1 }\)
\({ I }_{ 2 }=\frac { 1 }{ 5 } \int { \frac { 2x }{ ({ x }^{ 2 }+1) } } dx\)
Let  \({ x }^{ 2 }+1=t,\Rightarrow 2xdx=dt\)
\(=\frac { 1 }{ 5 } \int { \frac { dt }{ t } =\frac { 1 }{ 5 } logt } \)
\(=\frac { 1 }{ 5 } log(1+{ x }^{ 2 })+{ C }_{ 2 }\)
\({ I }_{ 3 }=\frac { 1 }{ 5 } \int { \frac { 1 }{ { x }^{ 2 }+1 } dx } \)
\(=\frac { 1 }{ 5 } tan^{ -1 }x+{ C }_{ 3 }\)
\(\therefore I=\frac { 1 }{ 5 } \left[ 3log(x+2)+log(1+{ x }^{ 2 })+tan^{ -1 }x \right] +C\)

Let \(I=\int_{0}^{\pi / 2} \frac{x \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x\) 
Using \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\) we get
\(I=\int_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x\) 
\(\Rightarrow I=\int_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \cos x \sin x}{\cos ^{4} x+\sin ^{4} x} d x\) 
\(\left[\because \cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta \text { and } \sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta\right]\) 
On adding Eqs. (i) and (ii), we get
\(2 I=\frac{\pi}{2} \int_{0}^{\pi / 2} \frac{\cos x \sin x}{\sin ^{4} x+\cos ^{4} x} d x\) 
\(\Rightarrow I=\frac{\pi}{4} \int_{0}^{\pi / 2} \frac{\sin x \cos x}{\left(\sin ^{2} x\right)^{2}+\left(1-\sin ^{2} x\right)^{2}} d x\) \(\left[\because \cos ^{2} \theta=1-\sin ^{2} \theta\right]\) 
Now, put \(\sin ^{2} x=t \Rightarrow 2 \sin x \cos x d x=d t\) 
\(\Rightarrow \sin x \cos x d x=\frac{d t}{2}\) 
Lower limit When \(x=0, \text { then } t=\sin ^{2} 0=0\) 
Upper limit When \(x=\frac{\pi}{2}, \text { then } t=\sin ^{2} \frac{\pi}{2}=1\) 
Now, \(I=\frac{\pi}{4} \int_{0}^{1} \frac{1}{t^{2}+(1-t)^{2}} \frac{d t}{2}\) 
\(\Rightarrow I=\frac{\pi}{8} \int_{0}^{1} \frac{1}{t^{2}+\left(1+t^{2}-2 t\right)} d t\) 
\(\Rightarrow I=\frac{\pi}{8} \int_{0}^{1} \frac{1}{2 t^{2}-2 t+1} d t \Rightarrow I=\frac{\pi}{16} \int_{0}^{1} \frac{1}{t^{2}-t+\frac{1}{2}} d\) 
\(\Rightarrow I=\frac{\pi}{16} \int_{0}^{1} \frac{1}{t^{2}-t+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+\frac{1}{2}} d t\) 

Given, the circle \(x^{2}+y^{2}=32\) 
and the line y = x 
Let us find the point of intersection of Eqs. (i) and (ii
On substituting y = x in Eq. (i), we get
\(x^{2}+x^{2}=32\) 
\(\Rightarrow 2 x^{2}=32\) 
\(\Rightarrow x^{2}=16 \Rightarrow x=\pm 4\) 
Thus, the points of intersection are \(\begin{array}{c} (4,4) \text { and }(-4,-4) {[\because y=x]} \end{array}\) 
 
Clearly, the required area
= Area of shaded region OABO
\( =\int_{0}^{4} y(\text { line }) d x+\int_{4}^{4 \sqrt{2}} y(\text { circle }) d x\)
\(=\int_{0}^{4} x d x+\int_{4}^{4 \sqrt{2}} \sqrt{32-x^{2}} d x\)
\(\left[\because x^{2}+y^{2}=32 \Rightarrow y=\pm \sqrt{32-x^{2}} \text { and } y>0\right]\)
\(=\left[\frac{x^{2}}{2}\right]_{0}^{4}+\int_{4}^{4 \sqrt{2}} \sqrt{(4 \sqrt{2})^{2}-x^{2}} d x \) 
\(=\frac{1}{2}[16-0]+\frac{1}{2}\left[x \sqrt{(4 \sqrt{2})^{2}-x^{2}}+(4 \sqrt{2})^{2} \sin ^{-1}\left(\frac{x}{4 \sqrt{2}}\right)\right]_{4}^{4 \sqrt{2}}\) 
\(=8+\frac{1}{2}\left[\left(0+32 \sin ^{-1}(1)\right)-\left(4 \sqrt{32-16}+32 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)\right]\) 
\(=8+\frac{1}{2}\left[32 \sin ^{-1}(1)-16-32 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right]\) 
\(=8+\frac{1}{2}\left[32 \cdot \frac{\pi}{2}-16-32 \cdot \frac{\pi}{4}\right]\)
\(=8+\frac{1}{2}[16 \pi-16-8 \pi]=8+\frac{1}{2}[8 \pi-16]=8+4 \pi-8\)
\(=4 \pi \text { sq units }\)

Given curve is  4x² + y² = 36 
\(\begin{array}{ll} \therefore & \frac{4 x^2}{36}+\frac{y^2}{36}=1 \end{array}\)
\(\begin{array}{ll} \therefore & \frac{x^2}{9}+\frac{y^2}{36}=1 \end{array}\)              ...(i)
We know that the standard equation of ellipse is
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)                      ...(ii)
On comparing Eqs. (i) and (ii), we get
a² = 9 and b² = 36
\(\Rightarrow\) a = 3 and b = 6
Here, we see that a < b, so the vertical ellipse will be formed.

Now, required area = 4(Area of region OAB in first quadrant)
\(\begin{aligned} =4\left|\int_0^3 y d x\right| \end{aligned}\)
\(\begin{aligned} =\left|4 \int_0^3 2 \sqrt{9-x^2} d x\right| \end{aligned}\)
\(\begin{aligned} \left[\because \frac{y^2}{36}=1-\frac{x^2}{9} \Rightarrow y=2 \sqrt{9-x^2}\right] \end{aligned}\)
\(\begin{aligned} =8 \int_0^3 \sqrt{9-x^2} d x \end{aligned}\)
\(\begin{aligned} =8\left|\left[\frac{x}{2} \sqrt{9-x^2}+\frac{9}{2} \sin ^{-1}\left(\frac{x}{3}\right)\right]_0^3\right| \end{aligned}\)
\(\left[\because \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]\)
\(=8\left\{\left[\frac{3}{2} \sqrt{9-9}+\frac{9}{2} \sin ^{-1} \frac{3}{3}-0-\frac{9}{2} \sin ^{-1} 0\right]\right\}\)
\(\begin{aligned} =8\left|\left[0+\frac{9}{2} \times \frac{\pi}{2}-0\right]\right| \end{aligned}\)
\(\begin{aligned} =8 \times \frac{9 \pi}{4} \end{aligned}\)
\(=18 \pi\)
Hence, the required area is 18\(\pi\) sq units.

We have, maximise, Z = 60x + 40y             ...(i)
Subject to the constraints, x + 2y \(\leq 12\)     ...(ii)
\(\begin{aligned} 2 x+y & \leq 12 \end{aligned}\)                                   ...(iii)
\(\begin{aligned} 4 x+5 y & \geq 20 \end{aligned}\)                                 ...(iv)
and \(\begin{aligned} x, y & \geq 0 \end{aligned}\)                                     ...(v)
Table for  line x + 2y = 12 is

So, the line x + 2y = 12 is passing through the points (0, 6) and (12, 0).
On putting (0,0) in the inequality \(x+2 y \leq 12\) we get \(0+2(0) \leq 12 \Rightarrow 0 \leq 12\), which is true
So,the half plane is towards the origin.
Table for line 2x + y = 12 is 

So, the line 2x + y = 12 is passing through the points (0, 12) and (6, 0).
On putting (0, 0) in the inequality \(2 x+y \leq 12\) we get
\(2(0)+0 \leq 12 \Rightarrow 0 \leq 12\) , which is true.
So, the half plane is towards the origin.
Table for line 4x + 5y = 20 is

So, the line 4x + 5y = 20 is passing through the points (0,4) and (5, 0).
On putting (0, 0) in the inequality \(4 x+5 y \geq 20\), we get 4(0) + 5(0) \(\geq\) 20 \(\Rightarrow\) 0 > 20 which is not true.
So, the half plane is away from the origin.
Also, x, y \(\geq\) 0
So, the region lies in Ist quadrant.

On solving Eqs. x + 2y = 12 and 2x + y = 12, we get D(4, 4)
Clearly, the feasible region is ABCDEA.
The corner points of the feasible region are A(0, 4), B(5, 0), C(6, 0), D(4, 4) and E(0, 6).
The value of Z at corner points are given below.

The maximum value of Z is 400 at D(4, 4).

Let \(I=\int_{-\pi}^\pi(\cos a x-\sin b x)^2 d x\)
\(=\int_{-\pi}^\pi\left(\cos ^2 a x+\sin ^2 b x-2 \cos a x \sin b x\right) d x \)
\(=\int_{-\pi}^\pi\left(\cos ^2 a x+\sin ^2 b x\right) d x-2 \int_{-\pi}^\pi \cos a x \sin b x d x \)
\(=I_1-I_2\)
Now consider, 
\(I_1=  \int_{-\pi}^\pi\left(\cos ^2 a x+\sin ^2 b x\right) d x \) [be an even function] 
\(=  2 \int_0^\pi\left(\cos ^2 a x+\sin ^2 b x\right) d x\)
\(\left[\because \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x, \text { if } f(x) \text { is even }\right]\)
\(=2 \int_0^\pi\left(\frac{1+\cos 2 a x}{2}+\frac{1-\cos 2 b x}{2}\right) d x\)
\(=\int_0^\pi(1+\cos 2 a x+1-\cos 2 b x) d x =\int_0^\pi(2+\cos 2 a x-\cos 2 b x) d x\)
\( =\left(2 x+\frac{\sin 2 a x}{2 a}-\frac{\sin 2 b x}{2 b}\right)_0^\pi\)
\(=\left(2 \pi+\frac{\sin 2 a \pi}{2 a}-\frac{\sin 2 b \pi}{2 b}\right)-0\)
\(=2 \pi+\frac{\sin 2 a \pi}{2 a}-\frac{\sin 2 b \pi}{2 b}\)
and \(I_2=2 \int_{-\pi}^\pi(\cos a x \sin b x) d x\) [be an odd function]
\(=0\left[\begin{array}{cc} \because \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x \text {, if } f(x) \text { is even } \\ 0, \text { if } f(x) \text { is odd } \end{array}\right]\)
\(\therefore I=I_1-I_2=2 \pi+\frac{\sin 2 a \pi}{2 a}-\frac{\sin 2 b \pi}{2 b}\)

As \(\int { \frac { 1 }{ \sqrt { 4-{ 9x }^{ 2 } } } dx } \)
\(=\frac { 1 }{ 3 } \int { \frac { 1 }{ \sqrt { { \left( \frac { 2 }{ 3 } \right) }^{ 2 }-{ x }^{ 2 } } } dx } \)
\(=\frac { 1 }{ 3 } \int { \frac { 1 }{ \sqrt { { \left( \frac { 2 }{ 3 } \right) }^{ 2 }-{ x }^{ 2 } } } dx } \)
⇒ a = \(\frac32\)