Question

How do you calculate the antiderivative of $\dfrac{{\sin (2x)}}{{\cos (x)}}dx$?

Answer 1

We know that antiderivative means integration. We need to find the integration of $\dfrac{{\sin (2x)}}{{\cos (x)}}dx$. Here we have an indefinite integral. In the numerator we have sine double angle, we know the sine double angle formula that is $\sin (2x) = 2.\sin x.\cos x$. We substitute this in the given problem and then we integrate with respect to ‘x’.

Complete step-by-step solution:
Given $\int {\dfrac{{\sin (2x)}}{{\cos (x)}}dx}$.
We know $\sin (2x) = 2.\sin x.\cos x$.
The term inside the integral symbol is called the integrand.
Then the integrand becomes
$\dfrac{{\sin (2x)}}{{\cos (x)}} = \dfrac{{2.\sin x.\cos x}}{{\cos x}}$
Cancelling the cosine function we have,
$\dfrac{{\sin (2x)}}{{\cos (x)}} = 2.\sin x.$
Now applying the integration we have
$\int {\dfrac{{\sin (2x)}}{{\cos (x)}}dx} = \int {2.\sin x} .dx$
$= \int {2.\sin x} .dx$
Taking constant term outside the integral we have,
$= 2\int {\sin x} .dx$
Integrating we have,
$= - 2\cos x + c$
Thus we have
The antiderivative of $\dfrac{{\sin (2x)}}{{\cos (x)}}dx$ is $- 2\cos x + c$. Where ‘c’ is the integration constant.

Note: In the given above problem we have an indefinite integral, that is no upper and lower limit. Hence we add the integration constant ‘c’ after integrating. In a definite integral we will have an upper and lower limit, we don’t need to add integration constant in the case of definite integral. We have different integration rule:
The power rule: If we have a variable ‘x’ raised to a power ‘n’ then the integration is given by $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c}$.
The constant coefficient rule: if we have an indefinite integral of $K.f(x)$, where f(x) is some function and ‘K’ represent a constant then the integration is equal to the indefinite integral of f(x) multiplied by ‘K’. That is $\int {K.f(x)dx = c\int {f(x)dx} }$.
The sum rule: if we have to integrate functions that are the sum of several terms, then we need to integrate each term in the sum separately. That is
$\int {\left( {f(x) + g(x)} \right)dx = \int {f(x)dx} } + \int {g(x)dx}$
For the difference rule we have to integrate each term in the integrand separately.