Question

integrate

$\int \: \frac{1}{ { \cos}^{2} x(1 - { \tan}x) ^{2} } dx$
​

Answer 1

To integrate :

$\star \: \int \: \frac{1}{ { \cos}^{2} x {(1 - \tan \: x)}^{2} } dx$

As we know that :

$\star \boxed{\bold{\frac{1}{ \cos \theta} = \sec \theta}}$

$\implies \: \int \: \frac{ { \sec}^{2} x}{ {(1 - \tan \: x)}^{2} } dx$

Now ,

we use substitution method :

Let ,

$\star \: 1 - \tan \: x = t$

Differentiate w.r.t x both sides,

$\star \: - { \sec}^{2} x = \frac{dt}{dx} \\ \\ \star \: dx = \frac{dt}{ - { \sec}^{2} x}$

Put the values :

$\implies \int \: \frac{ { \sec}^{2}x }{ {t}^{2} } \frac{dt}{ - { \sec}^{2}x } dx \\ \\ \implies - \int \: {t}^{ - 2} dt \\ \ \\ \implies \: - ( \frac{ { t}^{ - 2 + 1} }{ - 2 + 1} ) + c \\ \\ \implies \: \frac{1}{t} + c \\ \\ \implies \: \frac{1}{1 - \tan \: x} + c$

Formula :

$\star \bold{ \red{ \frac{d}{dx} \tan \: x \: = { \sec}^{2} x}} \\ \\ \star \bold{ \red{ \int \: {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1 } + c}} \:$

Answer 2

To integrate :

\begin{lgathered}\star \: \int \: \frac{1}{ { \cos}^{2} x {(1 - \tan \: x)}^{2} } dx \\\end{lgathered}

⋆∫

cos

2

x(1−tanx)

2

1

dx

As we know that :

\begin{lgathered}\star \boxed{\bold{\frac{1}{ \cos \theta} = \sec \theta}} \\\end{lgathered}

⋆

cosθ

1

=secθ

\begin{lgathered}\implies \: \int \: \frac{ { \sec}^{2} x}{ {(1 - \tan \: x)}^{2} } dx \\\end{lgathered}

⟹∫

(1−tanx)

2

sec

2

x

dx

Now ,

we use substitution method :

Let ,

\star \: 1 - \tan \: x = t⋆1−tanx=t

Differentiate w.r.t x both sides,

\begin{lgathered}\star \: - { \sec}^{2} x = \frac{dt}{dx} \\ \\ \star \: dx = \frac{dt}{ - { \sec}^{2} x}\end{lgathered}

⋆−sec

2

x=

dx

dt

⋆dx=

−sec

2

x

dt

Put the values :

\begin{lgathered}\implies \int \: \frac{ { \sec}^{2}x }{ {t}^{2} } \frac{dt}{ - { \sec}^{2}x } dx \\ \\ \implies - \int \: {t}^{ - 2} dt \\ \ \\ \implies \: - ( \frac{ { t}^{ - 2 + 1} }{ - 2 + 1} ) + c \\ \\ \implies \: \frac{1}{t} + c \\ \\ \implies \: \frac{1}{1 - \tan \: x} + c\end{lgathered}

⟹∫

t

2

sec

2

x

−sec

2

x

dt

dx

⟹−∫t

−2

dt

⟹−(

−2+1

t

−2+1

)+c

⟹

t

1

+c

⟹

1−tanx

1

+c

Formula :

\begin{lgathered}\star \bold{ \red{ \frac{d}{dx} \tan \: x \: = { \sec}^{2} x}} \\ \\ \star \bold{ \red{ \int \: {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1 } + c}} \:\end{lgathered}

⋆

dx

d

tanx=sec

2

x

⋆∫x

n

dx=

n+1

x

n+1

+c