Courses
Courses for Kids
Free study material
Offline Centres
More

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

seo-qna
Last updated date: 03rd Mar 2024
Total views: 341.1k
Views today: 7.41k
Rank Predictor
Answer
VerifiedVerified
341.1k+ views
Hint: 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.
Trending doubts