Mathematics

# Evaluate: $\displaystyle \int \dfrac{dx}{1-\tan x}$

##### SOLUTION
Solution:
$I=\displaystyle \int \dfrac {dx}{1-\tan x}$

$=\displaystyle \int \dfrac {\cos x}{\cos -\sin x}dx$

$=\displaystyle \int \dfrac {2\cos x}{2(\cos x-\sin x)}dx$

$=\dfrac {1}{2}\displaystyle \int \dfrac {\cos x+\cos x+\sin x-\sin x}{\cos x-\sin x}dx$

$=\dfrac {1}{2}\displaystyle \int \left (\dfrac {\cos x-\sin x}{\cos x-\sin x}+\dfrac {\cos x+\sin x}{\cos x-\sin x}\right)dx$

$=\dfrac {1}{2} \displaystyle \int 1dx+1 \dfrac {\cos x+\sin x}{\cos x-\sin x}dx$

Let $\cos x -\sin x=t$

$\Rightarrow \ (-\sin x-\cos x)dx=dt$

$\Rightarrow \ -(\sin x+\cos x)dx=dt$

$\Rightarrow \ (\sin x+\cos x)dx=dt$

$I=\dfrac {1}{2} \left [x+\displaystyle \int {-\dfrac {dt}{t}}\right]$

$=\dfrac {1}{2}[x-\log |t|]+C$

$=\dfrac {1}{2}[x-\log |\cos x-\sin x|]+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Subjective Medium
Evaluate :
$\displaystyle \int (|x - 1| + |x - 2| + |x - 4|)dx$.

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
If $\int \left( x ^ { 6 } + 7 x ^ { 5 } + 6 x ^ { 4 } + 5 x ^ { 3 } + 4 x ^ { 2 } + 3 x + 1 \right) e ^ { x } d x$ is equal to $\sum _ { k = 1 } ^ { \infty } \beta x ^ { k } \cdot e ^ { x } + c$  (where C is constant of integration) then $( \alpha + \beta )$ is-
• A. $8$
• B. $9$
• C. $10$
• D. $7$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Evaluate: $\displaystyle \int_{-2}^{2}(x^{11}\cos x+e^{x})dx$
• A. $\sinh 2$
• B. $\displaystyle \frac{3}{2} \sinh 2$
• C. $\displaystyle \frac{\sinh2}{2}$
• D. $2\sinh 2$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard

$I:\displaystyle \int\frac{a\cos x+b\sin x}{c\cos x+d\sin x}dx=$$\displaystyle \frac{ac+bd}{c^{2}+d^{2}}x+\frac{ad-bc}{c^{2}+d^{2}}\log|c\cos x+d\sin x|+k$
$II:\displaystyle \int\frac{\cos x+\sin x}{\sin x-\cos x}dx=\log|\sin x-\cos x|+k$ Which of the following is true
• A. Only I
• B. Only II
• C. neither I nor II are true
• D. Both I and II

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$