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Question: Find the antiderivative (or integral) of the following functions by method of inspection. \[\begin...

Find the antiderivative (or integral) of the following functions by method of inspection.

& 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{aligned}$$
Explanation

Solution

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, ddx(cos2x)=2sin2x\dfrac{\text{d}}{\text{dx}}\left( \cos 2x \right)=-2\sin 2x
    sin2x=12ddx(cos2x)\Rightarrow \sin 2x=-\dfrac{1}{2}\dfrac{\text{d}}{\text{dx}}\left( \cos 2x \right)
    Since, 12-\dfrac{1}{2} is constant number
    sin2x=ddx(cos2x2)\Rightarrow \sin 2x=\dfrac{\text{d}}{\text{dx}}\left( -\dfrac{\cos 2x}{2} \right)
    Hence cos2x2-\dfrac{\cos 2x}{2} is anti-derivative of sin2x.

  2. cos3x
    Let f(x) = cos3x
    We know that, ddx(sin3x)=3cos3x\dfrac{\text{d}}{\text{dx}}\left( \sin 3x \right)=3\cos 3x
    cos3x=13ddx(sin3x)\Rightarrow \cos 3x=\dfrac{1}{3}\dfrac{\text{d}}{\text{dx}}\left( \sin 3x \right)
    Since, 13\dfrac{1}{3} is constant number
    cos3x=ddx(sin3x3)\Rightarrow \cos 3x=\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\sin 3x}{3} \right)
    Hence sin3x3\dfrac{\sin 3x}{3} is anti-derivative of cos3x.

  3. e2x{{\operatorname{e}}^{2x}}
    Let f(x) =e2x{{\operatorname{e}}^{2x}}
    We know that, ddx(e2x)=2e2x\dfrac{\text{d}}{\text{dx}}\left( {{\operatorname{e}}^{2x}} \right)=2{{\operatorname{e}}^{2x}}
    e2x=12ddx(e2x)\Rightarrow {{\operatorname{e}}^{2x}}=\dfrac{1}{2}\dfrac{\text{d}}{\text{dx}}\left( {{\operatorname{e}}^{2x}} \right)
    Since, 12\dfrac{1}{2} is constant number
    e2x=ddx(e2x2)\Rightarrow {{\operatorname{e}}^{2x}}=\dfrac{\text{d}}{\text{dx}}\left( \dfrac{{{\operatorname{e}}^{2x}}}{2} \right)
    Hence e2x2\dfrac{{{\operatorname{e}}^{2x}}}{2} is anti-derivative of e2x{{\operatorname{e}}^{2x}}.

  4. (ax+b)2{{\left( ax+b \right)}^{2}}
    Let f(x) = (ax+b)2{{\left( ax+b \right)}^{2}}
    We know that, ddx(xn)=nxn1\dfrac{\text{d}}{\text{dx}}\left( {{x}^{n}} \right)=n{{x}^{n-1}}
    ddx(ax+b)3=3a(ax+b)2\dfrac{\text{d}}{\text{dx}}{{\left( ax+b \right)}^{3}}=3a{{\left( ax+b \right)}^{2}}
    (ax+b)2=13addx(ax+b)3\Rightarrow {{\left( ax+b \right)}^{2}}=\dfrac{1}{3a}\dfrac{\text{d}}{\text{dx}}{{\left( ax+b \right)}^{3}}
    Since, 13a\dfrac{1}{3a} is constant number
    (ax+b)2=ddx((ax+b)33a)\Rightarrow {{\left( ax+b \right)}^{2}}=\dfrac{\text{d}}{\text{dx}}\left( \dfrac{{{\left( ax+b \right)}^{3}}}{3a} \right)
    Hence (ax+b)33a\dfrac{{{\left( ax+b \right)}^{3}}}{3a} is anti-derivative of (ax+b)2{{\left( ax+b \right)}^{2}}

  5. sin2x4e2x\sin 2x-4{{\operatorname{e}}^{2x}}
    Let f(x) =sin2x4e2x\sin 2x-4{{\operatorname{e}}^{2x}}
    We know that, ddx(eax)=aeax and ddx(cosax)=asinax\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
    ddx(e3x)=3e3x and ddx(cos2x)=2sin2x\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

e3x=13ddx(e3x) and sin2x=12ddx(cos2x)\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, -12 and 13\text{-}\dfrac{\text{1}}{\text{2}}\text{ and }\dfrac{\text{1}}{\text{3}} is constant number
e3x=ddx(e3x3)and sin2x=ddx(cos2x2)\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
sin2x4e3x=ddx(cos2x2)4ddx(e3x3)\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)
sin2x4e3x=ddx(cos2x24e3x3)\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 cos2x24e3x3-\dfrac{\cos 2x}{2}-\dfrac{4{{\operatorname{e}}^{3x}}}{3} is anti-derivative of sin2x4e3x\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 ddx(sinax)=acosax\dfrac{\text{d}}{\text{dx}}\left( \operatorname{sina}x \right)=a\operatorname{cosa}x, ddx(cosax)=asinax\dfrac{\text{d}}{\text{dx}}\left( \operatorname{cosa}x \right)=-asinax, ddx(eax)=eax\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.