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@Vinod

$\\\int \sqrt{\frac{\sin \left(x-a\right)}{\sin \left(x+a\right)}}dx\\rationalize\;it\\\int \sqrt{\frac{\sin \left(x-a\right)\sin \left(x-a\right)}{\sin \left(x+a\right)\sin \left(x-a\right)}}dx\\\int \frac{\sin \left(x-a\right)}{\sqrt{\sin \:\left(x+a\right)\sin \:\left(x-a\right)}}dx\\\because \sin \left(A+B\right)\sin \left(A-B\right)=\sin ^2A-\sin ^2B\\\int \frac{\sin x\:\cos a-\sin a\:\cos x}{\sqrt{\sin ^2x-\sin ^2a}}dx\\\cos a\int \frac{\sin x\:}{\sqrt{\sin ^2x-\sin ^2a}}dx\:-\sin a\int \:\frac{\:\cos \:x}{\sqrt{\sin \:^2x-\sin \:^2a}}dx\:\:\:$

$\\\cos a\int \frac{\sin x\:}{\sqrt{\sin ^2x-\sin ^2a}}dx\:\\\cos \:a\int \frac{\sin \:x\:}{\sqrt{1-\cos ^2x-1+\cos ^2a}}dx\:\\\Rightarrow \cos \:a\int \frac{\sin \:x\:}{\sqrt{\cos \:^2a-\cos ^2x}}dx\:\\put\;\cos x=t,and\;solve\\same\;method\;apply\;for\;\sin a\int \:\frac{\:\cos \:x}{\sqrt{\sin \:^2x-\sin \:^2a}}dx\:\:\:$

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Let          Integrating by parts    =>  =>                      So, option (3) is correct.     Option 1) -1 Option 2) 1 Option 3) Option 4)
Option 1) Option 2) Option 3) Option 4)
86 Views   |
As we have learned Integration of trigonometric function of power m -     - wherein for  use   ,                Option 1) This is incorrect Option 2) This is correct Option 3) This is incorrect Option 4) This is incorrect
75 Views   |
As we have learned Sum/difference rule for integration -  . . . -     Option 1) This is incorrect Option 2) This is incorrect Option 3) This is incorrect Option 4) This is correct
92 Views   |
As we have learned Rule for integration - Integration of differential of a function is the function itself .   - wherein where           988 Option 1) 0 This is incorrect Option 2) -1 This is incorrect Option 3) 1 This is correct Option 4) 2 This is incorrect
91 Views   |
As we have learned Rule for integration - Integration of differential of a function is the function itself .   - wherein where            Option 1) Option 2) Option 3) Option 4)
78 Views   |
As we have learned Result for integration by parts -   - wherein Put                 Option 1) Option 2) Option 3) Option 4)
111 Views   |
As we have learned Result for integration by parts -   - wherein Put             Option 1) This is incorrect Option 2) This is incorrect Option 3) This is correct Option 4) This is incorrect
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As we have learned Result for integration by parts -   - wherein Put               Option 1) Option 2) Option 3) Option 4) none of these
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As we learnt Type of Integration by perfect square - Integrals in the form of      - wherein Working rule :   Find  and  by comparing  and       Option 1) Option 2) Option 3) Option 4)
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As we learnt Type of Integration by perfect square - Integrals in the form of      - wherein Working rule :   Find  and  by comparing  and       Diff. both sides, we get Comparing like terms on both sides, we get 3 = 2a – 3b, 2 = 3a + 2b Þ        . Option 1) Option 2) Option 3) Option 4) none of these
87 Views   |
As we learnt Type of Integration by perfect square - Integrals in the form of      - wherein Working rule :   Find  and  by comparing  and          Option 1) Option 2) Option 3) Option 4) none
127 Views   |
As we learnt in  NEWTON LEIBNITZ THEOREM - -    f (2) = 6 ; f ' (2) = 1/48     Option 1) 36 This is incorrect option Option 2) 24 This is incorrect option Option 3) 18 This is correct option Option 4) 12 This is incorrect option
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As learnt in concept Properties of Definite Integration - For periodic function Let Period (T) then   - wherein Where is periodic function with period T and n is any integer.     => Option 1) 20 This is incorrect Option 2) 8 This is incorrect Option 3) 10 This is incorrect Option 4) 18. This is correct
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As learnt in concept Integration by substitution - The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.     - wherein Since  all variables must be converted into single variable ,     = => If      Option 1) Option 2) Option 3) Option 4)
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