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General Solution of Trigonometric Equation

∫[cosx/x-sinx...

Question

$\int [\frac{c o s x}{x} - sin x log x] d x =$

A

(log x) s i n x + c

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B

(log x) (c o s x) + c

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C

2 (log x) (c o s x) + c

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D

- (log x) s i n x + c

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Solution

The correct option is A $(log x) (c o s x) + c$
$\int [\frac{c o s x}{x} - s i n x l o g x] d x$
$\int \frac{c o s x}{x} d x - \int s i n x l o g x d x$
$c o s x \int \frac{1}{x} d x + \int s i n x l o g x . d x$ $- \int s i n x l o g x d x .$
$= (c o s x) (l o g x) + c$
Hence, $\int [\frac{c o s x}{x} - (s i n x) (l o g x)] d x = (c o s x) (l o g x) + c$

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Similar questions

Q. Find the domains of the following functions :

f (x) = {log}_{x} sin x

\int sin x log (sec x + tan x) d x = f (x) + x + c

f (x) =

Q. How many of the following are matched correctly?

INTEGRAND	INTEGRAL
$\int x^{n} d x$	$\frac{x^{n}}{n + 1} + C$
$\int \frac{1}{x} d x$	$ln (x \| + C$
$\int e^{x} d x$	$e^{x} + C$
$\int a^{x} d x$	$\frac{a^{x}}{ln a} + C$
$\int cos x d x$	$sin x + C$
$\int sin x d x$	$cos x + C$
$\int sec x tan x d x$	$sec x + C$
$\int cosec x cot x d x$	$- cosec x + C$
$\int {sec}^{2} x d x$	$tan x + C$
$\int {cosec}^{2} x d x$	$cot x + C$

I = \int \frac{{sin}^{2} x + sin x}{1 + sin x + cos x} d x

J = \int \frac{{cos}^{2} x + cos x}{1 + sin x + cos x} d x

c

is the constant of integration.

\begin{matrix} Function & Integral \begin{matrix} (a) I \end{matrix} & (p) \frac{1}{2} (x - sin x - cos x) + c \begin{matrix} (b) J \end{matrix} & (q) \frac{1}{2} (x + sin x + cos x) + c \begin{matrix} (c) I + J \end{matrix} & (r) x + c \begin{matrix} (d) I - J \end{matrix} & (s) c - cos x - sin x \begin{matrix} \end{matrix} & (t) c + cos x + sin x \begin{matrix} \end{matrix} & (u) - \frac{1}{2} (x + sin x + cos x + c) \end{matrix}

Then the value of

\frac{d (I + J)}{d x}

x = \sqrt{2}

\int \frac{d x}{sin x + tan x}

\frac{1}{m} ln | \frac{(c o s x - 1)}{(c o s x + 1)} | - \frac{1}{2} \cdot \frac{1}{(c o s x + 1)} + C

m =

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General Solution of Trigonometric Equation

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