The value of integral ∫e^x (1-x/1+x²)² dx=___?

In order to find the value of given integral \[\int e^{x}\left ( \frac{1-x}{1+x^{2}} \right )^{2} dx\].

For that consider property of integral: $\displaystyle \int e^{x}\left [ f(x)+{f}'(x) \right ] dx=e^{x} f(x)+C$ 

Hence

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

Here take $\displaystyle f(x)=\frac{1}{(1+x^{2})}\; \mathbf{and}\; {f}'(x)= \frac{-2x}{(1+x^{2})^{2}}$ and now by using property of integrals 

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

Hence $\displaystyle \begin{align*} \int e^{x}\left [ \frac{1-x }{1+x^{2}}\right ]^{2} dx &= \frac{e^{x}}{1+x^{2}}+C \end{align*}$ 

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

 

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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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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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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50+ Integration formulas