Courses
Courses for Kids
Free study material
Offline Centres
More Last updated date: 06th Dec 2023
Total views: 282k
Views today: 8.82k

# Evaluate $\int {\dfrac{{cosx - \sin x}}{{\cos x + \sin x}} \times (2 + sin2x)dx = ?}$ Verified
282k+ views
Hint: In order to determine the given integral function. Here, whenever we are trying to solve any integral, we have to first look for constant terms, and after that we have to check for any formulas which are suitable to apply for the given. Here we have applied different Trigonometric formulas like: $1 + \sin 2x = {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x$
and basic algebraic formulas like:
${(a + b)^2} = {a^2} + {b^2} + 2ab$
$(a - b)(a + b) = {a^2} - {b^2}$
So, we are going to use the above mentioned formulas to achieve our solution.

In this problem,
We are given the function. $\int {\dfrac{{cosx - \sin x}}{{\cos x + \sin x}} \times (2 + sin2x)dx}$
In the above equation, we can see there is a constant which is number2, let us bring it out of integral.,
$= 2\int {\dfrac{{cosx - \sin x}}{{\cos x + \sin x}} \times (1 + \sin 2x)dx}$ ----------(1)
Now, if we look at the above equation, there is $\sin 2x$ term, there is a formula for this term which is as follows:
$\Rightarrow \sin 2x = 2\sin x\cos x$
Now, substitute this identity in the term $1 + \sin 2x$ of equation (1).
$\Rightarrow 1 + \sin 2x = 1 + 2\sin x\cos x$
$\Rightarrow 1 + \sin 2x = {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x$
Now, let us use the above formula in the place of $1 + \sin 2x$ , then we get;
$= 2\int {\dfrac{{cosx - \sin x}}{{\cos x + \sin x}} \times ({{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x)dx}$
Now, the above term $({\sin ^2}x + {\cos ^2}x + 2\sin x\cos x)$ is in the form of ${a^2} + {b^2} + 2ab$ which is equal to ${(a + b)^2}$ . So, let us convert it into that form, then we get;
$= 2\int {\dfrac{{(cosx - \sin x) \times {{(\cos x + sinx)}^2}}}{{\cos x + \sin x}}dx}$
$= 2\int {(cosx - \sin x) \times (\cos x + \sin x)dx}$
Now, the above equation is in the form of $(a - b)(a + b)$ which is equal to $({a^2} - {b^2})$ . Now, we will also convert it to the same form, then we will get,
$= 2\int {({{\cos }^2}x - {{\sin }^2}x)dx}$
Now, the above term is in the form of $({\cos ^2}x - {\sin ^2}x)$ , which is the formula of $\cos 2x$
$= 2\int {(\cos 2x)dx}$
Now, we know the formula for $\cos 2x$ , which is $\dfrac{{\sin 2x}}{2} + c$ , where c is constant, then we get:
$= 2 \times \dfrac{{\sin 2x}}{2} + c$
Where both number 2 get cancelled and the final answer is:
$= \sin 2x + c$
So, the correct answer is “ $\sin 2x + c$ ”.

Note: We note that in this problem, we have used basic Trigonometric formulas and basic algebraic formulas to achieve our solution which are very-very important.
Trigonometry is one of the branches of mathematics that uses trigonometric ratios to find the angles and missing sides of a triangle. The trigonometric functions are the trigonometric ratios of a triangle. The trigonometric functions are sine, cosine, sectant, cosecant, tangent and cotangent.