The integration by substitution rule applies if the integrand consists two parts (functions) in which one function is derivative of other (in some form). The two general structure of this form is \( \int f(x)^n.f(x)' dx\) or \( \int \frac{f(x)' dx}{f(x)^n} \)
That is, if a one part is the derivative of another via the chain rule, we let \(u\) to denote a likely candidate for the inner part as \(f(x)\), then translate the given function so that it is written entirely in terms of \(u\), with no \(x\) remaining in the expression.
If we can integrate this new function of \(u\), then the anti-derivative of the original function is obtained by replacing \(u\) by the equivalent expression in \(x\).
It can be further classified into three sub categories
Solution
Here, Derivative of \(\dfrac{d}{dx}x^2=2x\), therefore, we let \( u=x^2\)
Now \( \frac{du}{dx}= 2x\) or \(du=2xdx\)
Hence, we have \( \int (x^2-5)^7 2xdx\) or
\( \int (u-5)^7 du\) or
\( \int (u-5)^7 du\) or
\( \frac{(u-5)^8}{8}+c\) or
\( \frac{(x^2-5)^8}{8}+c\)
Solution
Here, Derivative of \(x^2\) is 2x, therefore, we let \( u=x^2\)
Now \(\frac{du}{dx}= 2x\) or du=2xdx
Since we have exactly 2xdx in the original integral, we can replace it by du, hence, we have \(\int 2x (\cos x^2) dx\) or
\(\int (\cos u) du \) or
\(\int 2x (\cos x^2) dx=\sin u+c\) or
\(\int 2x (\cos x^2) dx=\sin (x^2)+c\)
\(\int (ax+b)^ndx\), assuming that a and b are constants, \( a\ne 0\), and n is a positive integer.
Solution
We let u=ax+b so du=adx or dx=\(\frac{du}{a}\). Then \(\int (ax+b)^ndx\)
or
\(\int u^n \frac{du}{a}\) or
\(\frac{1}{a} \int u^n du\) or
\(\frac{1}{a} \frac{u^{n+1}}{n+1}+c\) or
\(\frac{1}{a(n+1)}(ax+b)^{n+1}+c\)
\( \int \sin(ax+b)dx\), assuming that a and b are constants and \(a\ne 0\).
Solution
We let u=ax+b so du=adx or dx=\( \frac{du}{a}\)
Then
\( \int \sin(ax+b)dx\) or
\( \int \sin(u) \frac{du}{a}\) or
\( \frac{1}{a} (-\cos u)+c\) or
\( -\frac{1}{a} \cos (ax+b)+c \)
Solution
Here, integrand consists products of the sine, sine with odd exponent, so it can be integrated by using substitution of trigonometric identities. Thus, \( \int \sin^5xdx\) or
\( \int \sin x \sin^4xdx\) or
\( \int \sin x (1- \cos^2x)^2 dx\)
Now, we use \( u=\cos x\) then \(du=-\sin x dx\)
Hence, we have \( \int \sin^5xdx\) or
\(\int (1-u^2)^2 (-du)\) or
\(-\int (1-2u^2+u^4) du\) or
\(\int \sin^5xdx=-\int (1-2u^2+u^4) du\) or
\(-\left (u-\frac{2u^3}{3}+\frac{u^5}{5}+c \right )\) or
\(-\left (\cos x-\frac{2\cos^3 x}{3}+\frac{\cos^5x}{5}+c \right )\)
Solution
Here,integrand consists products of the sine and cosine, sine with even exponent, so it can be integrated by using substitution of trigonometric identities. Thus, \(\int \sin^2x \cos^2 dx\)
or
\( \int \frac{1}{8} (1-\cos 4x) dx \) or
\( \frac{1}{8}\int (1-\cos 4x) dx\) or
\( \frac{1}{8}\left (x-\frac{\sin 4x}{4} \right )+c \)
Solution
यस प्रश्नमा \(x-a=u \) मानौ , then \( dx=du\)
Now, the integral is \( \int \frac{\sin(x+a)}{\sin(x-a)} dx \)
or\( \int \frac{\sin(a+u+a)}{\sin(u)} du \)
or\( \int \frac{\sin(u+2a)}{\sin(u)} du \)
or\( \int \frac{\sin(u) \cos(2a)+\cos(u) \sin(2a)}{\sin(u)} du \)
or\( \int (\cos(2a)+\cot(u) \sin(2a) )du \)
or\( u \cos(2a)+ \log (\sin(u)) \sin(2a) +c \)
or\( (x-a) \cos(2a)+ \log (\sin(x-a)) \sin(2a) +c \)
Trigonometric Substitutions
This type of substitution is usually indicated when the function of integrate contains a polynomial expression to use the fundamental trigonometric identity as \(\cos^2x+\sin^2x =1\) or \(\cos^2 x=1-\sin^2 x \) \(\sec^2x-\tan^2x=1 \) or \( \sec^2x=1+\tan^2x \) \(\sec^2x-\tan^2x=1 \) or \( \tan^2x=\sec^2x-1\)
Therefore
if it contains \( 1-x^2\), we substitute \(x=\sin u\)
if it contains \(1+x^2\) we substitute \(x=\tan u\) and
if it contains \(x^2-1\), we substitute \(x=\sec u\).
Sometimes, we need something a bit different to handle constants other than one. Thus, key points are as follows
Put \( x=a\sin u \) for integration involving \( a^2-x^2\)
Put \( x=a\tan u \) for integration involving \( a^2+x^2\)
Put \( x=a\sec u \) for integration involving \( x^2-a^2 \)
Solution
This question is in the form \(1-x^2\), so we substitute, \(x=\sin u\), then \( dx=\cos u du\), thus we have \(\int \sqrt{1-x^2} dx\) or
\( \int \sqrt{1-\sin^2 u} \cos u du \) or
\( \int \cos^2 u du\) or
\( \frac{1+\cos 2 u}{2} du\) or
\( \frac{u}{2}+\frac{\sin 2 u}{4} +c \)(A)
Here \(x=\sin u \) or
\( u=\sin^{-1}x \)(1)
Next \(\sin 2u =2 \sin u \cos u \) or
\( \sin 2u=2 \sin \sin^{-1}x \sqrt{1-\sin ^2 u} \) or
\( \sin 2u=2 x \sqrt{1-x^2} \)(2)
Thus, using (1) and (2),the solution (A) is \( \frac{u}{2}+\frac{\sin 2 u}{4} +c \) or
\( \frac{\sin^{-1} x}{2}+\frac{2x \sqrt{1-x^2}}{4} +c \)
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