Mathematics

# Find $\int { \dfrac { 1 }{ 1+tanx } dx }$.

##### SOLUTION
$\begin{array}{l} \displaystyle \int{ \dfrac { { dx } }{ { 1+\tan x } } } \\ =\displaystyle \int{ \dfrac { { \cos{x}dx } }{ { \cos x+\sin x } } } \\ =\dfrac { 1 }{ 2 } \displaystyle \int { \dfrac { { \left( { \sin x+\cos x } \right) +\left( { \cos x-\sin x } \right) } }{ { \left( { \sin x+\cos x } \right) } } } dx \\ =\dfrac { x }{ 2 } +\dfrac { 1 }{ 2 } \displaystyle \int { \dfrac { { \left( { \cos x-\sin x } \right) } }{ { \sin x+\cos x } } } dx \\ \sin x+\cos x=t \\ =\dfrac { x }{ 2 } +\dfrac { 1 }{ 2 } \displaystyle \int { \dfrac { { dt } }{ t } } \\ I=\dfrac { x }{ 2 } +\dfrac { 1 }{ 2 } \log \left( { \sin x+\cos x } \right) +C \end{array}$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Subjective Hard
Evaluate the following integral:
$\displaystyle \int { \cfrac { { x }^{ 2 }-1 }{ { x }^{ 2 }+4 } } dx\quad$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \lim_{n\rightarrow \infty} \dfrac {\sqrt {1} + \sqrt {2} + ..... + \sqrt {n - 1}}{n\sqrt {n}} = 0$
• A. $\dfrac {1}{2}$
• B. $\dfrac {1}{4}$
• C. $0$ (zero)
• D. $\dfrac {1}{3}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate:
$\displaystyle \int \dfrac{5 \cos^{3}x+6 \sin^{3}x}{2 \sin^{2}x \cos^{2} x}dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 One Word Medium
$\displaystyle \int_{0}^{\pi }x\log \sin x= \frac{\pi ^{k}}{m}\log \frac{1}{n}.$.Find $k+m+n$ ?

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$