Question

Find the antiderivative (or integral) of the following functions by method of inspection.
\[\begin{align}
  & 1)\text{ }\sin 2x. \\
 & 2)\text{ }\cos 3x \\
 & 3)\text{ }{{\operatorname{e}}^{2x}} \\
 & 4)\text{ }{{\left( ax+b \right)}^{2}} \\
 & 5)\text{ }\sin 2x-4{{\operatorname{e}}^{3x}} \\
\end{align}\]

Answer 1

Hint: In this we will the anti-derivatives of the given function by method of inspection.
Method of inspection: In the method of inspection we find the function F(x) which has derivative f(x) with respect to x.

Complete step by step answer:
1) sin2x
Let f(x) = sin2x
We know that, $\dfrac{\text{d}}{\text{dx}}\left( \cos 2x \right)=-2\sin 2x$
$\Rightarrow \sin 2x=-\dfrac{1}{2}\dfrac{\text{d}}{\text{dx}}\left( \cos 2x \right)$
Since, $-\dfrac{1}{2}$ is constant number
$\Rightarrow \sin 2x=\dfrac{\text{d}}{\text{dx}}\left( -\dfrac{\cos 2x}{2} \right)$
Hence $-\dfrac{\cos 2x}{2}$ is anti-derivative of sin2x.
2) cos3x
Let f(x) = cos3x
We know that, $\dfrac{\text{d}}{\text{dx}}\left( \sin 3x \right)=3\cos 3x$
$\Rightarrow \cos 3x=\dfrac{1}{3}\dfrac{\text{d}}{\text{dx}}\left( \sin 3x \right)$
Since, $\dfrac{1}{3}$ is constant number
$\Rightarrow \cos 3x=\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\sin 3x}{3} \right)$
Hence $\dfrac{\sin 3x}{3}$ is anti-derivative of cos3x.
3) \[{{\operatorname{e}}^{2x}}\]
Let f(x) =\[{{\operatorname{e}}^{2x}}\]
We know that, \[\dfrac{\text{d}}{\text{dx}}\left( {{\operatorname{e}}^{2x}} \right)=2{{\operatorname{e}}^{2x}}\]
$\Rightarrow {{\operatorname{e}}^{2x}}=\dfrac{1}{2}\dfrac{\text{d}}{\text{dx}}\left( {{\operatorname{e}}^{2x}} \right)$
Since, $\dfrac{1}{2}$ is constant number
\[\Rightarrow {{\operatorname{e}}^{2x}}=\dfrac{\text{d}}{\text{dx}}\left( \dfrac{{{\operatorname{e}}^{2x}}}{2} \right)\]
Hence $\dfrac{{{\operatorname{e}}^{2x}}}{2}$ is anti-derivative of \[{{\operatorname{e}}^{2x}}\].
4) \[{{\left( ax+b \right)}^{2}}\]
Let f(x) = \[{{\left( ax+b \right)}^{2}}\]
We know that, $\dfrac{\text{d}}{\text{dx}}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
\[\dfrac{\text{d}}{\text{dx}}{{\left( ax+b \right)}^{3}}=3a{{\left( ax+b \right)}^{2}}\]
$\Rightarrow {{\left( ax+b \right)}^{2}}=\dfrac{1}{3a}\dfrac{\text{d}}{\text{dx}}{{\left( ax+b \right)}^{3}}$
Since, $\dfrac{1}{3a}$ is constant number
$\Rightarrow {{\left( ax+b \right)}^{2}}=\dfrac{\text{d}}{\text{dx}}\left( \dfrac{{{\left( ax+b \right)}^{3}}}{3a} \right)$
Hence $\dfrac{{{\left( ax+b \right)}^{3}}}{3a}$ is anti-derivative of ${{\left( ax+b \right)}^{2}}$

5) \[\sin 2x-4{{\operatorname{e}}^{2x}}\]
Let f(x) =\[\sin 2x-4{{\operatorname{e}}^{2x}}\]
We know that, \[\dfrac{\text{d}}{\text{dx}}\left( {{\operatorname{e}}^{ax}} \right)=a{{\operatorname{e}}^{ax}}\text{ and }\dfrac{\text{d}}{\text{dx}}\left( \cos ax \right)=-a\sin ax\]
\[\dfrac{\text{d}}{\text{dx}}\left( {{\operatorname{e}}^{3x}} \right)=3{{\operatorname{e}}^{3x}}\text{ and }\dfrac{\text{d}}{\text{dx}}\left( \cos 2x \right)=-2\sin 2x\]

$\Rightarrow {{\operatorname{e}}^{3x}}=\dfrac{1}{3}\dfrac{\text{d}}{\text{dx}}\left( {{\operatorname{e}}^{3x}} \right)\text{ and }\sin 2x=-\dfrac{1}{2}\dfrac{\text{d}}{\text{dx}}\left( \cos 2x \right)$
Since, $\text{-}\dfrac{\text{1}}{\text{2}}\text{ and }\dfrac{\text{1}}{\text{3}}$ is constant number
\[\Rightarrow {{\operatorname{e}}^{3x}}=\dfrac{\text{d}}{\text{dx}}\left( \dfrac{{{\operatorname{e}}^{3x}}}{3} \right)\text{and }\sin 2x=\dfrac{\text{d}}{\text{dx}}\left( -\dfrac{\cos 2x}{2} \right)\]
Multiplying exponential function by 4 and subtracting from sine function, we get
\[\Rightarrow \sin 2x-4{{\operatorname{e}}^{3x}}=\dfrac{\text{d}}{\text{dx}}\left( -\dfrac{\cos 2x}{2} \right)-4\dfrac{\text{d}}{\text{dx}}\left( \dfrac{{{\operatorname{e}}^{3x}}}{3} \right)\]
\[\Rightarrow \sin 2x-4{{\operatorname{e}}^{3x}}=\dfrac{\text{d}}{\text{dx}}\left( -\dfrac{\cos 2x}{2}-\dfrac{4{{\operatorname{e}}^{3x}}}{3} \right)\]
Hence \[-\dfrac{\cos 2x}{2}-\dfrac{4{{\operatorname{e}}^{3x}}}{3}\] is anti-derivative of \[\sin 2x-4{{\operatorname{e}}^{3x}}\].

Note: In this to find the anti-derivatives of the given function by method of inspection one should know the derivatives of all functions. Always remember that $\dfrac{\text{d}}{\text{dx}}\left( \operatorname{sina}x \right)=a\operatorname{cosa}x$, $\dfrac{\text{d}}{\text{dx}}\left( \operatorname{cosa}x \right)=-asinax$, $\dfrac{\text{d}}{\text{dx}}\left( {{e}^{ax}} \right)={{e}^{ax}}$. Always remember that derivative is opposite of integration. Try not to make any calculation errors.