Question

Evaluate \[\int {\dfrac{{2 - 3\sin x}}{{{{\cos }^2}x}}dx} \].

Answer 1

Hint: This is a simple integration question. To solve the above problem we will first separate terms and convert it into simpler terms. Then taking constant terms out of the integration of simpler terms and then doing integration we will get the answer.

Complete step-by-step answer:
\[\int {\dfrac{{2 - 3\sin x}}{{{{\cos }^2}x}}dx} \]
Separating the terms of numerator with respect to denominator we get,
\[ = \int {\left( {\dfrac{2}{{{{\cos }^2}x}} - \dfrac{{3\sin x}}{{{{\cos }^2}x}}} \right)} dx\]
As we know that the trigonometric function $ \dfrac{1}{{\cos x}} = \sec x $ , hence substituting it in the above and expanding it we get,
\[ = \int {\left( {2{{\sec }^2}x - \dfrac{{3\sin x}}{{\cos x}} \times \dfrac{1}{{\cos x}}} \right)} dx\]
As we know that the trigonometric function $ \dfrac{1}{{\cos x}} = \sec x $ and $ \dfrac{{\sin x}}{{\cos x}} = \tan x $ , then substituting them above equation we get,
 \[ = \int {\left( {2{{\sec }^2}x - 3\tan x.\sec x} \right)} dx\]
Separating the terms and taking integration separately we get,
\[ = \int {(2{{\sec }^2}x)} dx - \int {(3\tan x.\sec x)} dx\]
Taking constant terms of each part out of the integration,
\[ = 2\int {({{\sec }^2}x)} dx - 3\int {(\tan x.\sec x)} dx\]
As we know that,
 $ \int {{{\sec }^2}xdx = \tan x + C} $
And $ \int {(\tan x.\sec x)dx = \sec x + C} $
Hence substituting these formulae in the above equation we get,
\[ = 2\tan x - 3\sec x + C\]
Hence, the answer is \[2\tan x - 3\sec x + C\], where C stands for constant.

Note: Proof for $ \int {{{\sec }^2}xdx = \tan x + C} $ .
L.H.S- $ \int {{{\sec }^2}xdx} = I $
Multiplying and dividing tan x we get,
 $ I = \int {\dfrac{{{{\sec }^2}x.\tan x}}{{\tan x}}} dx $ ………………. (1)
Let, $ \sec x = t $
Differentiating it we get, $ dt = \sec x.\tan xdx $
Again, $ \tan x = \sqrt {{{\sec }^2}x - 1} = \sqrt {{t^2} - 1} $
Putting these values in equation 1,
 $ I = \int {\dfrac{t}{{\sqrt {{t^2} - 1} }}} dt $
Again let, $ {t^2} - 1 = u $
Differentiating it, $ 2tdt = du $
 $ \therefore $ $ I = \dfrac{1}{2}\int {{u^{ - \dfrac{1}{2}}}} du $
Integrating it we get,
 $ I = \dfrac{1}{2}\dfrac{{\sqrt u }}{{\dfrac{1}{2}}} = \sqrt {{t^2} - 1} = \tan x + C $
Hence, L.H.S = R.H.S.
Similarly we can prove $ \int {(\tan x.\sec x)dx = \sec x + C} $
Integration is the technique of finding a function g(x) from its derivative dg(x), which is equal to a given function f(x).
The derivative of an integral of a function is that original function or we can say differentiation undoes the result of integration.
You should practice more questions of derivative and integration.
To solve this type of question you should remember all the formulas, properties and rules of trigonometry, inverse trigonometry, derivative and integration.