Exams Joint Entrance Examination Mathematics Indefinite Integrals Question 28 Question 28 of 416 Hard NUMERICAL
If ∫ csc 5 x d x = α cot x csc x ( csc 2 x + 3 2 ) + β log x ∣ tan x 2 ∣ + C \int {\csc }^{5}xdx=\alpha \cot x\csc x({\csc }^{2}x+\frac{3}{2})+\beta {\log }_{x}|\tan \frac{x}{2}|+C ∫ csc 5 x d x = α cot x csc x ( csc 2 x + 2 3 ) + β log x ∣ tan 2 x ∣ + C
where α , β ∈ R \alpha ,\beta \in R α , β ∈ R and C C C is the constant of integration, then the value of 8 ( α + β ) 8(\alpha +\beta ) 8 ( α + β ) equals \_\_\_\_\_\_\_\_\_.
Answer: 1
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I = ∫ ( csc x ) 5 d x = ∫ ( csc x ) 3 ( csc x ) 2 d x = ( csc x ) 3 ∫ csc 2 x d x − ∫ ( d d x ( csc x ) 3 ∫ csc 2 x d x ) d x \begin{gathered} I=\int (\csc x{)}^{5}dx=\int (\csc x{)}^{3}(\csc x{)}^{2}dx \\ =(\csc x{)}^{3}\int {\csc }^{2}xdx- \\ \int (\frac{d}{dx}(\csc x{)}^{3}\int {\csc }^{2}xdx)dx \end{gathered} I = ∫ ( csc x ) 5 d x = ∫ ( csc x ) 3 ( csc x ) 2 d x = ( csc x ) 3 ∫ csc 2 x d x − ∫ ( d x d ( csc x ) 3 ∫ csc 2 x d x ) d x
= − cot x ( csc x ) 3 − ∫ 3 csc 2 x ⋅ ( − csc x cot x ) ( − cot x ) d x = − cot x ( csc x ) 3 − ∫ 3 csc 3 x cot 2 x d x = − cot x ( csc x ) 3 − 3 ∫ ( csc x ) 3 ( csc 2 x − 1 ) d x I = − cot x ( csc x ) 3 − 3 I + 3 ∫ ( csc x ) 3 d x 4 I = − cot x ( csc x ) 3 + 3 ∫ ( csc x ) 3 d x = − cot x ( csc x ) 3 + 3 I 1 I 1 = ∫ csc x ⋅ csc 2 x d x = csc x ( − cot x ) − ∫ ( − csc x cot x ) ( − cot x ) d x \begin{gathered} =-\cot x(\csc x{)}^{3}-\int 3{\csc }^{2}x\cdot (-\csc x\cot x)(-\cot x)dx \\ =-\cot x(\csc x{)}^{3}-\int 3{\csc }^{3}x{\cot }^{2}xdx \\ =-\cot x(\csc x{)}^{3}-3\int (\csc x{)}^{3}({\csc }^{2}x-1)dx \\ I=-\cot x(\csc x{)}^{3}-3I+3\int (\csc x{)}^{3}dx \\ 4I=-\cot x(\csc x{)}^{3}+3\int (\csc x{)}^{3}dx \\ =-\cot x(\csc x{)}^{3}+3{I}_{1} \\ {I}_{1}=\int \csc x\cdot {\csc }^{2}xdx=\csc x(-\cot x)- \\ \, \int (-\csc x\cot x)(-\cot x)dx \end{gathered} = − cot x ( csc x ) 3 − ∫ 3 csc 2 x ⋅ ( − csc x cot x ) ( − cot x ) d x = − cot x ( csc x ) 3 − ∫ 3 csc 3 x cot 2 x d x = − cot x ( csc x ) 3 − 3 ∫ ( csc x ) 3 ( csc 2 x − 1 ) d x I = − cot x ( csc x ) 3 − 3 I + 3 ∫ ( csc x ) 3 d x 4 I = − cot x ( csc x ) 3 + 3 ∫ ( csc x ) 3 d x = − cot x ( csc x ) 3 + 3 I 1 I 1 = ∫ csc x ⋅ csc 2 x d x = csc x ( − cot x ) − ∫ ( − csc x cot x ) ( − cot x ) d x
I 1 = − csc x cot x − ∫ csc x ( csc 2 x − 1 ) d x I 1 = − csc x cot x − I + ∫ csc x d x 2 l 1 = − csc x cot x + ln ∣ tan x 2 ∣ ⇒ 4 l = − cot x ( csc x ) 3 − 3 2 csc x cot x + 3 2 ln ∣ tan x 2 ∣ + C \begin{gathered} {I}_{1}=-\csc x\cot x-\int \csc x({\csc }^{2}x-1)dx \\ {I}_{1}=-\csc x\cot x-I+\int \csc xdx \\ 2{l}_{1}=-\csc x\cot x+\ln |\tan \frac{x}{2}| \\ \Rightarrow \, 4l=-\cot x(\csc x{)}^{3}-\frac{3}{2}\csc x\cot x+\frac{3}{2}\ln |\tan \frac{x}{2}|+C \end{gathered} I 1 = − csc x cot x − ∫ csc x ( csc 2 x − 1 ) d x I 1 = − csc x cot x − I + ∫ csc x d x 2 l 1 = − csc x cot x + ln ∣ tan 2 x ∣ ⇒ 4 l = − cot x ( csc x ) 3 − 2 3 csc x cot x + 2 3 ln ∣ tan 2 x ∣ + C
I = − 1 4 csc x cot x ( csc 2 x + 3 2 ) + 3 8 ln ∣ tan x 2 ∣ + C ⇒ α = − 1 4 , β = 3 8 ⇒ 8 ( α + β ) = 1 \begin{gathered} I=-\frac{1}{4}\csc x\cot x({\csc }^{2}x+\frac{3}{2})+\frac{3}{8}\ln |\tan \frac{x}{2}|+C \\ \Rightarrow \alpha =-\frac{1}{4},\beta =\frac{3}{8}\Rightarrow 8(\alpha +\beta )=1 \end{gathered} I = − 4 1 csc x cot x ( csc 2 x + 2 3 ) + 8 3 ln ∣ tan 2 x ∣ + C ⇒ α = − 4 1 , β = 8 3 ⇒ 8 ( α + β ) = 1 Answer the question to unlock