Question

How do you find the antiderivative of \[\int {\dfrac{{\sin x}}{{{{\cos }^3}x}}dx} ?\]

Answer 1

Hint: In order to solve this integral first we will assume \[\cos x = t\] and differentiate it and transfer the given integral in terms of \[t\] . And then we will use the formula as, \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c} \] to solve the given integral. And finally, we will substitute the value of \[t\] to get the required result.

Complete step by step answer:
We have to find the antiderivative of \[\int {\dfrac{{\sin x}}{{{{\cos }^3}x}}dx} \]
Let us consider the given integral as,
\[I = \int {\dfrac{{\sin x}}{{{{\cos }^3}x}}dx} {\text{ }} - - - \left( i \right)\]
Now let us assume
\[\cos x = t{\text{ }} - - - \left( {ii} \right)\]
As we know that
\[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]
So, by differentiating both the sides of equation \[\left( {ii} \right)\] w.r.t \[x\] we get
\[ - \sin x = \dfrac{{dt}}{{dx}}\]
On multiplying by \[dx\] both sides, we get
\[ - \sin xdx = dt{\text{ }}\]
On multiplying with negative sign both the sides, we get
\[ \Rightarrow \sin xdx = - dt{\text{ }} - - - \left( {iii} \right)\]
Now substituting the value from equation \[\left( {ii} \right)\] and equation \[\left( {iii} \right)\] in equation \[\left( i \right)\] we get
\[I = \int {\dfrac{{ - dt}}{{{t^3}}}} \]
Now we know that
\[\dfrac{1}{{{a^n}}} = {a^{ - n}}\]
Therefore, we get
\[I = \int { - {t^{ - 3}}dt} \]
As we know that
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c} \]
Therefore, we get
\[I = \dfrac{{ - {t^{ - 3 + 1}}}}{{ - 3 + 1}} + c\]
On solving the numerator and denominator, we get
\[I = \dfrac{{ - {t^{ - 2}}}}{{ - 2}} + c\]
On cancelling the negative sign, we get
\[I = \dfrac{{{t^{ - 2}}}}{2} + c\]
We know that
\[{a^{ - n}} = \dfrac{1}{{{a^n}}}\]
Therefore, from the above equation, we get
\[I = \dfrac{1}{{2{t^2}}} + c\]
Now using equation \[\left( {ii} \right)\] substitute the value of \[t\]
Therefore, we get
\[I = \dfrac{1}{{2{{\cos }^2}x}} + c\]
Hence, the antiderivative of \[\int {\dfrac{{\sin x}}{{{{\cos }^3}x}}dx} \] is \[\dfrac{1}{{2{{\cos }^2}x}} + c\]

Note:
Antiderivative is another name of the inverse derivative, or the indefinite integral. Always remember while calculating antiderivatives never forget to add constant \[c\] in the final result. But students should know that constant will be used only if we have indefinite integral.