Find the value of integral ∫x³/(x+1) dx ?

 

Consider the integral

 \[\int \frac{x^{3}}{(x+1)}\; dx = \int \frac{x^{3}+1 -1}{(x+1)}\; dx\]

We have the algebraic identity: $\displaystyle x^{3}+1= (x+1)(x^{2}-x+1)$ .

Hence 

\[\begin{align*} \int \frac{x^{3}+1 -1}{(x+1)}\; dx&=\int \left (\frac{(x^{3}+1)}{x+1} - \frac{1}{(x+1)} \right )\; dx \\ &=\int \frac{x^{3}+1}{x+1}\; dx - \int\frac{1}{x+1} \; dx \\ &=\int \frac{(x+1)(x^{2}-x+1)}{x+1}\; dx -\int \frac{1}{x+1}\; dx\\ &=\int (x^{2}-x+1)\; dx -\int \frac{1}{x+1}\; dx \; \; (\mathbf{Cancelling \; x+1})\\ &= \frac{x^{3}}{3}-\frac{x^{2}}{2}+x- \log \left |x+1 \right |+C\\ \end{align*}\]

And \[\begin{align*} \int \frac{x^{3}+1}{(x+1)}\; dx &= \frac{x^{3}}{3}-\frac{x^{2}}{2}+x- \log \left |x+1 \right |+C\\ \end{align*}\]

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Find the value of the integral $\displaystyle \int \frac{x+\sin x}{1+\cos x} dx$ =?

 

To find the value of the integral ,consider the following trigonometric formulas:

$\displaystyle \cos 2x= 2\cos ^{2} x-1 \Rightarrow \cos x=2\cos ^{2} \frac{x}{2} -1\Rightarrow 1+\cos x=2\cos ^{2} \frac{x}{2}\\ \sin 2x= 2 \sin x \cos x\Rightarrow \sin x= 2 \sin \frac{x}{2} \cos \frac{x}{2}$ 

Hence

  

\[\begin{align*} \int \frac{x+\sin x}{1+\cos x}\; dx&= \int \left (\frac{x}{1+\cos x}\; + \frac{\sin x}{1+\cos x}\; \right ) dx\\ &=\int \left (\frac{x}{2\cos^{2} \frac{x}{2}}\; + \frac{2\sin \frac{x}{2}\cos \frac{x}2{}}{2\cos^{2} \frac{x}{2}}\; \right ) dx \\ &= \int \left (\frac{1}{2} x \sec ^{2} \frac{x}{2}\; + \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\; \right ) dx \\ &= \frac{1}{2}\int x\sec ^{2} \frac{x}{2} \; dx +\int \tan \frac{x}{2} dx\\ &=\frac{1}{2}\left (x\int \sec ^{2} \frac{x}{2} \; dx -\int \left (\frac{\mathrm{d} x}{\mathrm{d} x}\int \sec ^{2}\frac{x}{2}\; dx\right ) dx \right ) +\int \tan \frac{x}{2} dx \\ &= \frac{1}{2}\left (x\; \frac{\tan \frac{x}{2}}{1/2}-\int \frac{\tan \frac{x}{2}}{1/2}dx \right )+\int \tan \frac{x}{2}dx \\&=x\tan \frac{x}{2}+C-\int \tan \frac{x}{2}dx+\int \tan \frac{x}{2}dx\\ &=x\tan \frac{x}{2}+C \end{align*}\]

\[\therefore \int \frac{(x+\sin x)}{1+\cos x} \; dx= x \; \tan \frac{x}{2} +C\]

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If ∫x³ dx/(√(1+x²) = a (1+ x²)³/² +b√(1+x²)+ C, then a and b is given by:$\\ \mathbf{\;A)\; a=\frac{1}{3} , b=1.\; B)\; a=-\frac{1}{3}, b=1\; \\ C)\; a=-\frac{1}{3}, b=-1\; D)\; a=\frac{1}{3},b=-1} $

 

We have 

$\displaystyle \int \frac{x^{3}}{\sqrt{(1+x^{2})}} \; dx= a(1+x^{2})^{\frac{3}{2}}+b\sqrt{(1+x^{2})}+C \cdots (1)$ 

In-order  to find value of a and b , at first we have to evaluate the integral $\displaystyle \int \frac{x^{3}}{\sqrt{(1+x^{2})}} \; dx$ .

Now

 $\displaystyle \int \frac{x^{3}}{\sqrt{(1+x^{2})}} \; dx=\int \frac{x^{2}\;.\: x}{\sqrt{(1+x^{2})}} \; dx$ 

Take 

$\displaystyle 1+x^{2}=u\Rightarrow 2x=\frac{\mathrm{d} u}{\mathrm{d} x}\mathbf{(Differentiating \; w.r.t \; x)}\\  \Rightarrow x \; dx=\frac{du}{2}$ 

Now by substituting $\displaystyle 1+x^{2}=u$  we will get

\[\begin{align*} \int \frac{x^{3}}{\sqrt{(1+x^{2})}} \; dx &= \int \frac{u-1}{\sqrt{u}} \frac{du}{2}\\ &= \frac{1}{2}\int \frac{u}{\sqrt{u}}- \frac{1}{\sqrt{u}} \; du\\ &=\frac{1}{2}[\int \frac{\sqrt{u}}{1}\; du-\int \frac{1}{\sqrt{u}}\; du] \\ &= \frac{1}{2}[\int u^{\frac{1}{2}}\; du-\int u^{-\frac{1}{2}}\; du]\\ &=\frac{1}{2}[\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}]+C \\ &=\frac{1}{2}[\frac{u^{3/2}}{3/2}-\frac{u^{1/2}}{1/2}] +C \\&= \frac{u^{3/2}}{3}-u^{1/2}+C \end{align*}\]

Now by providing value $\displaystyle u=1+x^{2}$  we will get:

$\displaystyle \int\frac{ x^{3}}{\sqrt{(1+x^{2})}} \; dx=\frac{1}{3}\left (1+x^{2} \right )^{\frac{3}{2}}-\left ( 1+x^{2} \right )^{\frac{1}{2}} +C \cdots(2)$ 

Comparing (1) and (2) we will get

$\displaystyle \mathbf{a=1/3\; and \; b=-1}$ 

Hence correct option :D

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$\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi }{4}} \frac{dx}{1+\cos 2x}=\\ a) \;\; 1 \; \; b)\;\; 2\; \; c)\; \; 3\; \; d)\; \; 4$

 

Let $\displaystyle f(x)=\frac{1}{1+\cos 2x}\\$ 

By the properties of definite integral we have

$\mathbf{\int_{-a}^{a} f(x) dx} =\left\{\begin{matrix} 2\int_{0}^{a} f(x)dx \; \; if\; \textbf{f is even}, f(-x)=f(x)\\ 0 \;\; \; \; \; \; \; \; \;\; \; \; \; \; \; \: if \: \textbf{ f is odd},f(-x)=-f(x) \end{matrix}\right.$ 

Here $\displaystyle f(-x)=f(x)$ 

$\displaystyle f(-x)=\frac{1}{1+\cos (-2x)}=\frac{1}{1+\cos 2x}=f(x) \; (\because \cos (-x)=\cos x)$ 

Hence 

$\displaystyle f(x)=\frac{1}{1+\cos 2x}$   is an even function

\[\therefore \int_{-\frac{\pi }{4}}^{\frac{\pi }{4}} \frac{1}{1+\cos 2x} \; dx=2\int_{0}^{\frac{\pi }{4}} \frac{1}{1+\cos 2x} \; dx\]

And

\[\begin{align*} \int_{-\frac{\pi }{4}}^{\frac{\pi }{4}} \frac{1}{1+\cos 2x} \; dx &=2\int_{0}^{\frac{\pi }{4}} \frac{1}{1+2\cos^{2} x-1} \; dx \; \; (\because \cos 2x=2 \cos ^{2} x-1) \\ &= 2\int_{0}^{\frac{\pi }{4}} \frac{1}{2\cos^{2} x} \; dx\; \\ &= \int_{0}^{\frac{\pi }{4}} \frac{1}{\cos^{2} x} \; dx \; (by\; \; cancelling \; 2's)\\ &=\int_{0}^{\frac{\pi }{4}}\sec^{2} x\; dx \\ &= \left [ \tan x \right ]_{0}^{\frac{\pi }{4}}\\ &=\tan \frac{\pi }{4}-\tan 0 \\ &=1 \end{align*}\]

Hence $\displaystyle \int_{-\frac{\pi }{4}}^{\frac{\pi }{4}} \frac{1}{1+\cos 2x} \; dx =1$ 

And option a) 1 is the correct answer

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$\displaystyle \int_{0}^{\pi /2} \sqrt{(1- \sin 2x)}=\\ a)2\sqrt{2} \; \; \; b)2(\sqrt{2}+1)\; \; \; c) 2 \; \; \; d)2(\sqrt{2}-1)$

 

Method 1

In order to find solution consider the properties of definite integral

Property1: 

$\displaystyle \int_{0}^{2a}f(x) \; dx =\begin{cases} 2\int_{0}^{a}f(x)\; dx & \text{ if } f(2a-x)=f(x) \\ 0 & \text{ if } f(2a-x)=-f(x) \end{cases}$ 

Property 2: 

$\displaystyle \int_{0}^{a} f(x) \; dx =\int_{0}^{a} f(a-x)\; dx$ 

Now consider the property 1, take $f(x)=\sqrt{1-\sin 2x}$ , $\displaystyle a= \frac{\pi }{4}$  and

$f(2a-x)=f(2\frac{\pi }{4} -x)=f(\frac{\pi }{2}-x)$ .

Here 

\[\begin{align*} f(2a-x) &=f(\frac{\pi }{2}-x ) \\ &= \sqrt{1-\sin 2(\frac{\pi }{2}-x)}\\ &= \sqrt{1-\sin (\pi -2x)}\\ &=\sqrt{1-\sin 2x} \; \; (\sin (\pi -x)=\sin x)\\ &= f(x) \end{align*}\]

Hence $\displaystyle \int_{0}^{2a}f(x)\; dx=2\int_{0}^{a}f(x)dx$ and

 \[\begin{align*} \int_{0}^{2\frac{\pi}{4}=\frac{\pi }{2}}\sqrt{1-\sin 2x}\; dx &=2\int_{0}^{ \frac{\pi}{4}}\sqrt{1-\sin 2x}\; dx \\ &=2\int_{0}^{ \frac{\pi}{4}}\sqrt{1-\sin 2(\frac{\pi }{4}-x)}\; dx \; (by\; property \; 2 )\\ &=2\int_{0}^{ \frac{\pi}{4}}\sqrt{1-\sin (\frac{\pi }{2}-2x)}\; dx \\&=2\int_{0}^{ \frac{\pi}{4}}\sqrt{1-\cos 2x}\; dx \: \; (\because \sin (\frac{\pi }{2}-2x)= \cos 2x)\\ &=2\int_{0}^{ \frac{\pi}{4}}\sqrt{2\sin ^{2} x}\; dx \; \; (\because 1-\cos 2x= 2\sin ^{2} x )\\ &=2\int_{0}^{ \frac{\pi}{4}}\sqrt{2}\sin x \; dx \; \; (\because \sqrt{\sin ^{2}x}=\sin x\; for\; 0\leq x\leq \frac{\pi }{4})\\ &=2\sqrt{2}\left [ -\cos x \right ]_{0}^{\frac{\pi }{4}} \\ &=2\sqrt{2} \left [ -\cos \frac{\pi }{4}+\cos 0 \right ]\\ &= 2\sqrt{2}[-\frac{1}{\sqrt{2}}+1]\\&= 2[-1+\sqrt{2}]\\&= 2[\sqrt{2}-1] \end{align*}\]

Hence correct answer:(d)

Method 2

In order to find solution consider the definite integral

\[\begin{align*} \int_{0}^{\frac{\pi }{2}} \sqrt{(1-\sin 2x)}\: dx &=\int_{0}^{\frac{\pi }{2}} \sqrt{(\sin ^{2} x+\cos ^{2} x-\sin 2x)} \; dx \; \; (\because 1=\sin ^{2} x+\cos ^{2} x)\\ &=\int_{0}^{\frac{\pi }{2}} \sqrt{(\sin ^{2} x+\cos ^{2} x-2\sin x\cos x)}\; dx \; \; (\sin 2x=2\sin x\cos x)\\ &=\int_{0}^{\frac{\pi }{2}} \sqrt{(\sin x-\cos x)^{2}}\; dx \\ &= \int_{0}^{\frac{\pi }{2}} \left |(\sin x-\cos x) \right |\; dx \end{align*}\]

Here we have value of x ranges from  $0\leq x\leq \frac{\pi }{2}$ .

So we can examine that :

When $\displaystyle 0\leq x\leq \frac{\pi }{4} \; \; [\sin x -\cos x] \; \mathbf{is \; negative }.$ and

When $\displaystyle \frac{\pi }{4}\leq x\leq \frac{\pi }{2} \; \; [\sin x -\cos x] \; \mathbf{is \; positive }.$ 

Hence

\[\begin{align*} \int_{0}^{\frac{\pi }{2}}\left | \sin x - \cos x \right | dx&=\int_{0}^{\frac{\pi }{4}}-(\sin x - \cos x ) dx+\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}(\sin x - \cos x ) dx \\ &=-\left [ (-\cos x -\sin x) \right ]_{0}^{\frac{\pi }{4}}+[-\cos x - \sin x ]_{\frac{\pi }{4}}^{\frac{\pi }{2}} \\ &=-[(-\cos \frac{\pi }{4}-\sin \frac{\pi }{4})-(-\cos 0-\sin 0)]+[(-\cos \frac{\pi }{2}-\sin \frac{\pi }{2})-(-\cos \frac{\pi }{4}-\sin \frac{\pi }{4})] \\ &=- [(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})-(-1-0)]+[(-0-1)-(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})]\\ &=-[-\frac{2}{\sqrt{2}}+1]+[-1+\frac{2}{\sqrt{2}}]\\ &=-2+2\frac{2}{\sqrt{2}} \end{align*}\]

\[\therefore \int_{0}^{\frac{\pi }{2}}\sqrt{(1-\sin 2x)}=\int_{0}^{\frac{\pi }{2}}\left | \sin x - \cos x \right | dx =-2+2\frac{2}{\sqrt{2}}=-2+2\sqrt{2} \\ (\because 2=(\sqrt{2})^{2})\]

Hence 

\[\int_{0}^{\frac{\pi }{2}}\sqrt{(1-\sin 2x)} =2\sqrt{2}-2 =2(\sqrt{2}-1)\]

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$\int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x \; dx$ = ...

 

For finding the value of the integral $\int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x \; dx$  we are going to use substitution method.

i.e., Put  $\sin x=u$ 

Now by differentiating with respect to x we will get:

\[\begin{align*} \frac{\mathrm{d} (\sin x)}{\mathrm{\! d} x} &= \frac{\mathrm{d} u}{\mathrm{d} x} \Rightarrow \cos x = \frac{\mathrm{d} u}{\mathrm{d} x}\\ & \; \;\; \; \; \; \; \; \; \Rightarrow \cos x \; dx = du \end{align*}\]

Now by substituting  u instead of $\sin x$ we will get

\[\int e^{\sin x} \cos x \; dx = \int e^{u} \; du.\]

Next we have to apply limit for finding the value of integral.We have two methods:

Method1: Applying limit after integrating the function

\[\int e^{u} \; du. = e^{u} = e^{\sin x} (\because definite \; not \; using\; constant C) \]

\[\int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x\; dx. = \left [e^{\sin x} \right ]_{0}^{\frac{\pi }{2}}\]

\[\begin{align*} \int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x\; dx. &=\left [e^{\sin x} \right ]_{0}^{\frac{\pi }{2}} \\ &= e^{\sin \frac{\pi }{2}} - e^{0}\\ &= e^{1} - 1 \end{align*}\]

$\displaystyle \therefore \int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x\; dx. = e - 1$ 

Method2: Changing the limit before evaluating integral.

At first we have to find limits of integration before evaluating it.

i.e,We have to find value of u when $\displaystyle x=0 \; and \; x=\pi /2$ .

$x=0 \; \Rightarrow u= \sin 0=0 $ 

$\\ x=\pi /2 \; \Rightarrow u=\sin \frac{\pi }{2}=1$ 

Hence 

\[\int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x \; dx =\int_{0}^{1} e^{u} du\]

\[\int_{0}^{1} e^{u} \; du =\left [e^{u} \right ]_{0}^{1}=e^{1} - e^{0}=e-1\]

$\displaystyle \therefore \int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x\; dx. = e - 1$ 

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Find the value of the integral $\int \frac{(x+3)}{(x+4)^{2}} e^{x} \: dx$

 For finding the value of the integral $\int \frac{(x+3)}{(x+4)^{2}} e^{x} \: dx$  , consider the property of indefinite integral:

\[\int e^{x} [f(x) + {f}'(x)] dx= e^{x}f(x)+C\]

Consider the function:

\[\begin{align*} \left [\frac{x+3}{(x+4)^{2}} \right ] e^{x} & =\left [ \frac{x+3+1-1}{(x+4)^{2}} \right ]e^{x} \\ &= \left [ \frac{x+4}{(x+4)^{2}} - \frac{1}{(x+4)^{2}} \right ] e^{x}\\ &=\left [ \frac{1}{x+4} - \frac{1}{(x+4)^{2}}\right ] e^{x}\\ &= \left [ f(x)+{f}'(x) \right ] e^{x} \end{align*}\]

where 

\[f(x)=\frac{1}{x+4} \; and\; {f}'(x)=\frac{\mathrm{d}f }{\mathrm{d} x}=\frac{-1}{(x+4)^{2}}\]

So by the property of definite integral given above

\[\begin{align*} \int e^{x}\left [ f(x) +{f}'(x) \right ] dx &= \int e^{x}\left [ \frac{1}{(x+4)}-\frac{1}{(x+4)^{2}} \right ] dx\\ &= e^{x}\left [\frac{1}{x+4} \right ]+C\; \; \left (=e^{x}f(x)+C \right ) \end{align*}\]

\[\therefore \int e^{x}\left [\frac{x+3}{(x+4)^{2}} \right ] dx = e^{x}\left [\frac{1}{x+4} \right ]+C\]

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The value of $\int_{-\pi }^{\pi } \sin ^{3} x \cos ^{2} x \: dx=... ?$

 For finding the value of $\int_{-\pi }^{\pi } \sin ^{3} x \cos ^{2} x \: \mathbf{dx}$ :

Consider the property of definite integral 

$\mathbf{\int_{-a}^{a} f(x) dx} =\left\{\begin{matrix} 2\int_{0}^{a} f(x)dx \; \; if\; \textbf{f is even}, f(-x)=f(x)\\ 0 \;\; \; \; \; \; \; \; \;\; \; \; \; \; \; \: if \: \textbf{ f is odd},f(-x)=-f(x) \end{matrix}\right.$ 

Here we have $f(x)=\sin ^{3} x \cos ^{2} x$ 

And  \[\begin{align*} f(-x) &= \sin ^{3}(-x) \cos ^{2}(-x) \\ &= (\sin (\, -x))^{3}(\cos (\, -x))^{2} \\ &= (-\sin x)^{3} (\cos x)^{2} \; \; [\because \sin (-x)=-\sin x \; and\; \cos (-x)=\cos x]\\ &= -\sin ^{3} x \cos ^{2} x\\ &= -f(x) \end{align*}\]

Hence $f(x)=\sin ^{3} x \cos ^{2} x$  is a odd function.

And 

\[\int_{-\pi }^{\pi } f(x) \; dx = \int_{-\pi }^{\pi } \sin ^{3}x \cos ^{2} x\; dx=0\]

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If $\int_{0}^{a} 1/(1+4x^{2}) dx=\pi /8$ ,then a=?

We have  \[\int_{0}^{a} 1/(1+4x^{2}) dx=\pi /8\] and our aim is to find the value of a .

So in order find a ,at first we have solve the integral or to find the value of given integral.For that consider the formulae $\mathbf{\int \frac{1}{a^{2}+(bx)^{2}} dx= \frac{1}{ab}\tan ^{-1}\frac{bx}{a}} + C$ 

Now consider the integral 

\[\int_{0}^{a} 1/(1+4x^{2}) dx=\int_{0}^{a} 1/(1^2+(4x)^{2}) dx\]

Here a=1,b=2  now on integrating we will get 

\[\int_{0}^{a} \frac{1}{(1+4x^{2})} dx=\left [\frac{1}{1.2} \tan ^{-1} \frac{2x}{1} \right ]_{0}^{a} =\frac{1}{2}[\tan ^{-1} 2a - \tan ^{-1} 0] =\frac{1}{2}\tan ^{-1} 2a \; \: (\because \tan ^{-1} 0=0)\]

\[\therefore \int_{0}^{a} \frac{1}{(1+4x^{2})} dx=\frac{1}{2}\tan ^{-1} 2a \;\] 

Also we have  $\int_{0}^{a} 1/(1+4x^{2}) dx=\pi /8 $.

Now by using these two equations we can write

\[\begin{align*} \frac{1}{2}\tan ^{-1} 2a \; = \; \pi /8 &\Rightarrow \tan ^{-1} 2a= 2\frac{\pi }{8} =\frac{\pi }{4}\\ &\Rightarrow 2a=\tan \frac{\pi }{4} \\ &\Rightarrow 2a=1 \\ &\Rightarrow a=1/2 \end{align*}\]

Hence value of a=1/2.

 

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Find value of integral ∫(𝐬𝐢𝐧 𝐱)/(3+4𝐜𝐨𝐬² 𝐱 ) 𝐝𝐱

To find the value of this integral 

Consider the integral $\int \frac{sin x }{3+ 4 cos^2 x}$ we can rewrite it as

\[\int \frac{\textbf{sinx}}{{3+ 4 \cos^2 x}}=\int \frac{\sin x}{3+ (2\cos x)^2}\]

For finding the value put  $ 2\cos x=t$

By differentiating with respect to x then we will get -2 sin x  = dt/dx \[\Rightarrow \sin x dx = -dt/2\]

Hence

\[\int \frac{\textbf{sinx}}{{3+ 4 \cos^2 x}}=\int \frac{-dt}{2(\sqrt{3})^2+ (t)^2)}  \]

We have  the formulae $\int \mathbf{\frac{1}{a^2 + x^2}}dx = \frac{1}{a} \tan ^{-1} \frac{x}{a}+C$ .

Now by using that we can find value of the given integral 

i.e., 

 \[\int \frac{-dt}{2((\sqrt{3})^2 +t^2)}=\frac{-1}{2\sqrt{3}}\tan ^{-1}\frac{t}{\sqrt{3}}+C\] 

Now by substituting value of t we will get the value of integral as

\[\int \frac{\textbf{sinx}}{{3+ 4 \cos^2 x}} dx=\frac{-1}{2\sqrt{3}}\tan ^{-1}(\frac{2 \cos x}{\sqrt{3}})+C\]

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