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Answer 1

$\int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} } \frac{dx}{1 + \cos x}$

Now, Multiplying and Divide by

$1 - \cos x$

So,

$\int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} } \frac{dx}{1 + \cos x} \times \frac{1 - \cos x}{1 - \cos x}$

$= > \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} } \frac{ (1 - \cos x) dx}{(1 + \cos x)( 1 - \cos x) }$

$= > \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} } \frac{ (1 - \cos x) dx}{{(1 )}^{2} - {( \cos x)}^{2} }$

We know that ${1 - {( \cos x)}^{2} } = { \sin }^{2} x$

So,

$= > \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} } \frac{ (1 - \cos x) dx}{ { \sin }^{2} x}$

$= > \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} }( \frac{ (1 ) }{ { \sin }^{2} x} - \frac{ \cos x}{ { \sin}^{2} x} )dx$

$= > \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} }( \frac{ (1 ) }{ { \sin }^{2} x} - \frac{ \cos x}{ \sin x} \times \frac{1}{ \cos x } )dx$

As

$\underline{\overline{\mid{\bold{\blue{\mathcal{\frac{ (1 ) }{ { \sin }^{2} x} = { \cosec }^{2} x}}\mid}}}}$

,

$\underline{\overline{\mid{\bold{\blue{\mathcal{\frac{ \cos x}{ \sin x} = \cot x}}\mid}}}}$

and

$\underline{\overline{\mid{\bold{\blue{\mathcal{\frac{1}{ \cos x } = \cosec x}}\mid}}}}$

So, Substituting the values

$= > \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} }({ \cosec }^{2} x - \cot x \cosec x )dx$

$= >[ - \cot x + \cosec x ]^ \frac{3\pi}{4} _ \frac{\pi}{4}$

$= > ( 1 - \sqrt{2} ) - ( - 1 + \sqrt{2} )$

$= > 2$

Answer 2

Answer:

\int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} } \frac{dx}{1 + \cos x} ∫

4

π

4

3π

1+cosx

dx

Now, Multiplying and Divide by

1 - \cos x1−cosx

So,

\int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} } \frac{dx}{1 + \cos x} \times \frac{1 - \cos x}{1 - \cos x} ∫

4

π

4

3π

1+cosx

dx

×

1−cosx

1−cosx

= > \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} } \frac{ (1 - \cos x) dx}{(1 + \cos x)( 1 - \cos x) } =>∫

4

π

4

3π

(1+cosx)(1−cosx)

(1−cosx)dx

= > \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} } \frac{ (1 - \cos x) dx}{{(1 )}^{2} - {( \cos x)}^{2} } =>∫

4

π

4

3π

(1)

2

−(cosx)

2

(1−cosx)dx

We know that {1 - {( \cos x)}^{2} } = { \sin }^{2} x1−(cosx)

2

=sin

2

x

So,

= > \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} } \frac{ (1 - \cos x) dx}{ { \sin }^{2} x} =>∫

4

π

4

3π

sin

2

x

(1−cosx)dx

= > \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} }( \frac{ (1 ) }{ { \sin }^{2} x} - \frac{ \cos x}{ { \sin}^{2} x} )dx=>∫

4

π

4

3π

(

sin

2

x

(1)

−

sin

2

x

cosx

)dx

= > \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} }( \frac{ (1 ) }{ { \sin }^{2} x} - \frac{ \cos x}{ \sin x} \times \frac{1}{ \cos x } )dx=>∫

4

π

4

3π

(

sin

2

x

(1)

−

sinx

cosx

×

cosx

1

)dx

As

\underline{\overline{\mid{\bold{\blue{\mathcal{\frac{ (1 ) }{ { \sin }^{2} x} = { \cosec }^{2} x}}\mid}}}}

∣

sin

2

x

(1)

=cosec

2

x∣

,

\underline{\overline{\mid{\bold{\blue{\mathcal{\frac{ \cos x}{ \sin x} = \cot x}}\mid}}}}

∣

sinx

cosx

=cotx∣

and

\underline{\overline{\mid{\bold{\blue{\mathcal{\frac{1}{ \cos x } = \cosec x}}\mid}}}}

∣

cosx

1

=cosecx∣

So, Substituting the values

= > \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} }({ \cosec }^{2} x - \cot x \cosec x )dx=>∫

4

π

4

3π

(cosec

2

x−cotxcosecx)dx

= > [ - \cot x + \cosec x ]^ \frac{3\pi}{4} _ \frac{\pi}{4} =>[−cotx+cosecx]

4

π

4

3π

= > ( 1 - \sqrt{2} ) - ( - 1 + \sqrt{2} )=>(1−

2

)−(−1+

2

)

= > 2=>2

Explanation:

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Warm regards:Miss Chikchiki

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