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Question: Evaluate the integral \(\int {\dfrac{{{{\sec }^2}x - 7}}{{{{\sin }^7}x}}} dx = \) \(A)\left[ {\dfr...

Evaluate the integral ∫sec⁡2x−7sin⁡7xdx=\int {\dfrac{{{{\sec }^2}x - 7}}{{{{\sin }^7}x}}} dx =
A)[tan⁡xsin⁡7x]+cA)\left[ {\dfrac{{\tan x}}{{{{\sin }^7}x}}} \right] + c
B)[cos⁡xsin⁡7x]+cB)\left[ {\dfrac{{\cos x}}{{{{\sin }^7}x}}} \right] + c
C)[sin⁡xcos⁡7x]+cC)\left[ {\dfrac{{\sin x}}{{{{\cos }^7}x}}} \right] + c
D)[sin⁡xtan⁡7x]+cD)\left[ {\dfrac{{\sin x}}{{{{\tan }^7}x}}} \right] + c

Explanation

Solution

The given question consists of two different trigonometric functions in the numerator and the denominator. Therefore, we need to find out the integral value through the Integration By parts method. Since it is constant, we will separate it from the attached trigonometric function. Then, we apply by parts for the expression which contains only trigonometric functions.

Complete answer:
Let,
I=∫sec⁡2x−7sin⁡7xdxI = \int {\dfrac{{{{\sec }^2}x - 7}}{{{{\sin }^7}x}}} dx
Now, we can separate the constant and write the expression as,
I=∫sec⁡2xsin⁡7xdx−∫7sin⁡7xdxI = \int {\dfrac{{{{\sec }^2}x}}{{{{\sin }^7}x}}} dx - \int {\dfrac{7}{{{{\sin }^7}x}}} dx
We can write 1sin⁡x\dfrac{1}{{\sin x}} as cos⁡ecx\cos ecx
Now, we have the expression,
⇒I=∫cos⁡ec7x×sec⁡2xdx−∫7cos⁡ec7xdx\Rightarrow I = \int {\cos e{c^7}} x \times {\sec ^2}xdx - \int {7\cos e{c^7}} xdx
We apply integration by parts for the above expression ∫f(x)g(x)dx=f(x)∫g(x)dx−∫[ddxf(x)∫g(x)dx]dx\int {f\left( x \right)g\left( x \right)dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } } , we get,
⇒I=cos⁡ec7x∫sec⁡2xdx−∫[ddx[cos⁡ec7x]∫sec⁡2xdx]−∫7cos⁡ec7xdx\Rightarrow I = \cos e{c^7}x\int {{{\sec }^2}xdx} - \int {\left[ {\dfrac{d}{{dx}}\left[ {\cos e{c^7}x} \right]\int {{{\sec }^2}xdx} } \right]} - \int {7\cos e{c^7}} xdx
We know that ∫sec⁡2xdx=tan⁡x\int {{{\sec }^2}xdx} = \tan x and d(cos⁡ecx)dx=−cos⁡ecxcot⁡x\dfrac{{d\left( {\cos ecx} \right)}}{{dx}} = - \cos ecx\cot x.
⇒I=cos⁡ec7xtan⁡x−∫[7cos⁡ec6x(−cos⁡ecxcot⁡x)tan⁡x]−∫7cos⁡ec7xdx\Rightarrow I = \cos e{c^7}x\tan x - \int {\left[ {7\cos e{c^6}x\left( { - \cos ecx\cot x} \right)\tan x} \right]} - \int {7\cos e{c^7}} xdx
Now,
⇒I=cos⁡ec7xtan⁡x−∫[7cos⁡ec6x(−cos⁡ecxcot⁡x)tan⁡x]−∫7cos⁡ec7xdx\Rightarrow I = \cos e{c^7}x\tan x - \int {\left[ {7\cos e{c^6}x\left( { - \cos ecx\cot x} \right)\tan x} \right]} - \int {7\cos e{c^7}} xdx
Simplifying the expression, we get,
⇒I=cos⁡ec7xtan⁡x+∫[7cos⁡ec7xcot⁡xtan⁡x]−∫7cos⁡ec7xdx\Rightarrow I = \cos e{c^7}x\tan x + \int {\left[ {7\cos e{c^7}x\cot x\tan x} \right]} - \int {7\cos e{c^7}} xdx
Now, we know that tangent and cotangent are reciprocal functions. So, we get,
⇒I=cos⁡ec7xtan⁡x+∫7cos⁡ec7xdx−∫7cos⁡ec7xdx\Rightarrow I = \cos e{c^7}x\tan x + \int {7\cos e{c^7}xdx} - \int {7\cos e{c^7}} xdx
Now, cancelling the similar terms with opposite signs, we get,
Where, cos⁡ec7x\cos e{c^7}x can be expressed as 1sin⁡7x\dfrac{1}{{{{\sin }^7}x}}
Therefore, the final answer will be
I=tan⁡xsin⁡7x+cI = \dfrac{{\tan x}}{{{{\sin }^7}x}} + c
Hence, option (A) is the correct option.

Note:
The integration by parts method is useful when two functions are multiplied together and have no specific formula for simplification. If the terms are not in the multiplication form, we need to bring them down into that form using the basic formula. Only then we can use the formula and simplify.