Question

Integrate the following functions:
A) $\sin 7x.\cos 3x$
B) $\dfrac{{{{\sin }^4}2x + {{\cos }^4}2x}}{{{{\sin }^2}2x.{{\cos }^2}2x}}$

Answer 1

In this question, we have to find the integration of $(\sin 7x.\cos 3x)$. See clearly, the given expression is in product form. Change this in the sum of two sines. After doing so, use the form $\int {\left( {a + b} \right)dx = \int {a.dx + \int {b.dx} } }$
Take $10x$ and $4x$ as $t$ and $u$ respectively. Doing so, a standard form will generate. Just evaluate it and put back the values of $t$ and $u$. You will get the answer.

Complete step-by-step answer:
${ \int {\left( {\sin 7x.\cos 3x} \right)dx} \\\ = \dfrac{1}{2}\int {\left( {2\sin 7x.\cos 3x} \right)} dx \\\ = \dfrac{1}{2}\int {\left( {\sin 10x + \sin 4x} \right)dx} \\\ = \dfrac{1}{2}\left[ {\int {\left( {\sin 10x} \right)dx + \int {\left( {\sin 4x} \right)dx} } } \right].............(i) \\\ }$
(By applying formula of $\left[ {\because 2\sin A\cos B = \sin \left( {A + B} \right)\cos \left( {A + B} \right)} \right]$
Let us consider, value of $10x = t.............(ii)$
$\therefore$Now by differentiating both sides with respect to $x$, we get

\therefore \dfrac{d}{{dx}}\left( {10x} \right) = \dfrac{{dt}}{{dx}} \\\ \Rightarrow 10 = \dfrac{{dt}}{{dx}} \\\ \Rightarrow dt = 10dx \\\ } $$ Again, let’s consider the value of $4x$ be $u$. $$4x = u$$ $$\therefore $$ Again, by differentiating both sides with respect to $$x$$,we get $${ \therefore \dfrac{d}{{dx}}\left( {4x} \right) = \dfrac{d}{{dx}}\left( u \right) \\\ \therefore 4 = \dfrac{{du}}{{dx}} \\\ \Rightarrow du = 4dx \\\ } $$ Now putting the value of $$10x = t$$ and $$4x = u$$ in equation………………(i), we get $$ = \dfrac{1}{2}\left[ {\int {\sin t.dx} + \int {\sin u.dx} } \right]$$ $$ = \dfrac{1}{2}\left[ {\int {\sin t.} \dfrac{{dt}}{{10}} + \int {\sin u.\dfrac{{du}}{4}} } \right]$$ $$\left[ { \operatorname{Sin} ce,10x = t,x = \dfrac{t}{{10}} \\\ 4x = u \Rightarrow x = \dfrac{u}{4} \\\ } \right]$$ Now, we know by the formulas of integration, $$\int {\sin t.dt = - \cos t} $$ Therefore, we get here- $$ = \dfrac{1}{2}\left[ {\dfrac{{ - \cos t}}{{10}} - \dfrac{{\cos u}}{4}} \right] + c$$ $$ = - \dfrac{{\cos t}}{{20}} - \dfrac{{\cos u}}{8} + c$$ $$\left[ { \because \int {\sin t.dt = - \cos t} \\\ \int {\sin u.du = - \cos u} \\\ } \right]$$ Putting the values of $$t$$ and $$u$$ in the determined equation, we get the ultimate answer $$ = - \dfrac{{\cos 10x}}{{20}} - \dfrac{{\cos 4x}}{8} + c$$ **Note:** To solve the questions regarding integrations required knowledge of all the methods of solving integral form. In such a question, substitute the function and try to bring it in a standard form. Once you take out the standard form, then you can easily solve using formulas. This is the best way to solve such typical questions and also give much concentration on the product forms. If you found any product forms of trigonometry functions. Convert them into the sum of two trigonometric functions.