Question

Find the value of
$\int {\dfrac{1}{{\sin x\cos x\left( {{{\tan }^9}x + 1} \right)}}} dx = \\\$
A. $\log \left| {\dfrac{{{{\cos }^9}x}}{{{{\sin }^9}x + {{\cos }^9}x}}} \right| + c\\\$
B . $\dfrac{1}{9}\log \left| {\dfrac{{{{\sin }^9}x}}{{{{\sin }^9}x + {{\cos }^9}x}}} \right| + c\\\$
C. $\dfrac{1}{9}\log \left| {\dfrac{{{{\sin }^9}x + {{\cos }^9}x}}{{{{\sin }^9}x}}} \right| + c\\\$
D. $\dfrac{1}{9}\log \left| {\dfrac{{{{\cos }^9}x}}{{{{\sin }^9}x + {{\cos }^9}x}}} \right| + c\\\$

Answer 1

The given function is indefinite since there is no limit given. The indefinite integral of a function f is a differentiable function F whose derivative is equal to the original function f. The first fundamental theorem of calculus allows definite integrals to be computed in terms of indefinite integrals.

Complete step by step solution:
Let the given integral be $I$ such that:
$I = \int {\dfrac{1}{{\sin x\cos x\left( {{{\tan }^9}x + 1} \right)}}} dx - - (i)$
Now divide the numerator and the denominator of the given integral by ${\cos ^2}x$
$I = \int {\dfrac{{\left( {\dfrac{1}{{{{\cos }^2}x}}} \right)}}{{\left( {\dfrac{{\sin x\cos x\left( {{{\tan }^9}x + 1} \right)}}{{{{\cos }^2}x}}} \right)}}} dx - - (ii)$
We know that the relation $\sec x = \dfrac{1}{{\cos x}}$ holds true as the Reciprocal identities, and the relation $\tan x = \dfrac{{\sin x}}{{\cos x}}$ holds true as the Quotient identities.
Hence by using these trigonometric identities, we get integral in equation (ii) as
$I = \int {\dfrac{{{{\sec }^2}x}}{{\tan x\left( {{{\tan }^9}x + 1} \right)}}} dx - - (iii)$
Now let $\tan x = t - - - - (iv)$
Differentiating the equation (iv) with respect to x, we get
${\sec ^2}xdx = dt$
Substituting the value of $\tan x = t$and ${\sec ^2}xdx = dt$in integral I in the equation (iii), we get

$$I = \int {\dfrac{{{{\sec }^2}x}}{{\tan x\left( {{{\tan }^9}x + 1} \right)}}} dx \\\ = \int {\dfrac{{dt}}{{t\left( {{t^9} + 1} \right)}}} - - - - (v) \\\$$

Take ${t^9}$ common form the terms in the denominator of the equation (v) as:

$$I = \int {\dfrac{{dt}}{{t\left( {{t^9} + 1} \right)}}} \\\ = \int {\dfrac{{dt}}{{{t^{10}}\left( {1 + \dfrac{1}{{{t^9}}}} \right)}}} - - - - (vi) \\\$$

Let, $1 + \dfrac{1}{{{t^9}}} = v - - - - (vii)$
Differentiate the equation (vii) with respect to ‘t’ we get
$\dfrac{d}{{dt}}\left( {1 + \dfrac{1}{{{t^9}}}} \right) = \dfrac{{dv}}{{dt}} \\\ 0 - \dfrac{9}{{{t^{10}}}} = \dfrac{{dv}}{{dt}} \\\ \dfrac{{dv}}{{dt}} = - \dfrac{9}{{{t^{10}}}} \\\ dt = \dfrac{{ - {t^{10}}}}{9}dv \\\$
Now, substitute $dt = \dfrac{{ - {t^{10}}}}{9}dv$ and $1 + \dfrac{1}{{{t^9}}} = v$ in the equation (vi), we get

$$I = \int {\dfrac{{dt}}{{{t^{10}}\left( {1 + \dfrac{1}{{{t^9}}}} \right)}}} \\\ = \int {\dfrac{{\left( {\dfrac{{ - {t^{10}}}}{9}dv} \right)}}{{{t^{10}} \times v}}} \\\ = \dfrac{{ - 1}}{9}\int {\dfrac{{dv}}{v}} - - - - (viii) \\\$$

Integrating the terms in the equation (viii), we get

$$I = \dfrac{{ - 1}}{9}\int {\dfrac{{dv}}{v}} \\\ = \dfrac{{ - \log \left| v \right|}}{9} + c - - - - (ix) \\\$$

Now, back substituting the values from equation (vii) to the equation (ix), we get

$$I = \dfrac{{ - \log \left| v \right|}}{9} + c \\\ = \dfrac{{ - \log \left| {1 + \dfrac{1}{{{t^9}}}} \right|}}{9} + c - - - - (x) \\\$$

Again back substituting the values from the equation (iv) to the equation (x), we get

$$I = \dfrac{{ - \log \left| {1 + \dfrac{1}{{{t^9}}}} \right|}}{9} + c \\\ = \dfrac{{ - \log \left| {1 + \dfrac{1}{{{{\tan }^9}x}}} \right|}}{9} + c - - - - (xi) \\\$$

Replace $\tan x = \dfrac{{\sin x}}{{\cos x}}$ in the equation (xi), we get

$$I = \dfrac{{ - \log \left| {1 + \dfrac{1}{{{{\tan }^9}x}}} \right|}}{9} + c \\\ = \dfrac{{ - \log \left| {1 + \dfrac{{{{\cos }^9}x}}{{{{\sin }^9}x}}} \right|}}{9} + c \\\$$

On further simplifying:

$$I = \dfrac{{ - \log \left| {1 + \dfrac{{{{\cos }^9}x}}{{{{\sin }^9}x}}} \right|}}{9} + c \\\ = \dfrac{{ - \log \left| {\dfrac{{{{\sin }^9}x + {{\cos }^9}x}}{{{{\sin }^9}x}}} \right|}}{9} + c \\\ = \dfrac{{ - 1}}{9}\log \left| {\dfrac{{{{\sin }^9}x + {{\cos }^9}x}}{{{{\sin }^9}x}}} \right| + c \\\ = \dfrac{1}{9}\log \left| {{{\left( {\dfrac{{{{\sin }^9}x + {{\cos }^9}x}}{{{{\sin }^9}x}}} \right)}^{ - 1}}} \right| + c \\\ = \dfrac{1}{9}\log \left| {\dfrac{{{{\sin }^9}x}}{{{{\sin }^9}x + {{\cos }^9}x}}} \right| + c \\\$$

Hence, $\int {\dfrac{1}{{\sin x\cos x\left( {{{\tan }^9}x + 1} \right)}}} dx = \dfrac{1}{9}\log \left| {\dfrac{{{{\sin }^9}x}}{{{{\sin }^9}x + {{\cos }^9}x}}} \right| + c$

Option B is correct.

Important equations used:
$\cos 2\theta = 2{\cos ^2}\theta - 1$
$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
$\int {dx = x + c}$
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
$\sec x = \dfrac{1}{{\cos x}}$

Note: While substituting the real parameter of the question with the auxiliary parameter, one should be sure that it will not make the problem more complex. However, selecting an auxiliary parameter completely depends on the individual point of view.