Answer
Verified
414.6k+ views
Hint: To solve the above problem we have to use integration by parts formula and it is given by \[\int{u.v=u\int{v}}-\int{du\int{v}}\]. So consider u as \[\sin \left( \log x \right)\] and v as 1 and apply the integration by parts formula as stated above. The derivative of \[\sin t\] is \[\cos t\].
Complete step-by-step solution -
To find the \[\int{\sin \left( \log x \right)}dx\]
Take I as \[\int{\sin \left( \log x \right)}\times 1dx\]
\[I=\int{\sin \left( \log x \right)}\times 1dx\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . (1)
Apply integration by parts formula
\[I\]\[=\sin \left( \log x \right)\times x-\int{\cos (\log x)\times \dfrac{1}{x}}\times xdx\]
\[I\]\[=\sin \left( \log x \right)\times x-\int{\cos (\log x)\times 1}dx\]. . . . . . . . . . . . . . .(2)
\[I\]\[=\sin \left( \log x \right)\times x-\cos (\log x)\times x-\int{\sin \left( \log x \right)dx}\]
By equation(1) we substitute the value of $\int{\sin \left( \log x \right)dx}$ as $I$
\[I=\sin \left( \log x \right)\times x-\cos (\log x)\times x-I\]. . . . . . . . . . . . . . . . . . . . . . . . (3)
\[2I=x\left[ \sin \left( \log x \right)-\cos (\log x) \right]\]
\[I=\dfrac{x}{2}\left[ \sin \left( \log x \right)-\cos (\log x) \right]\]. . . . . . . . . . . . . . . . .. . (4)
The answer is option A.
Note: When they ask us to find the integration of a single function we can use normal integration formulas and if they ask us to find integration of product of two functions then we have to apply the integration by parts. If the two functions are different then we have to use the ILATE rule.
Complete step-by-step solution -
To find the \[\int{\sin \left( \log x \right)}dx\]
Take I as \[\int{\sin \left( \log x \right)}\times 1dx\]
\[I=\int{\sin \left( \log x \right)}\times 1dx\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . (1)
Apply integration by parts formula
\[I\]\[=\sin \left( \log x \right)\times x-\int{\cos (\log x)\times \dfrac{1}{x}}\times xdx\]
\[I\]\[=\sin \left( \log x \right)\times x-\int{\cos (\log x)\times 1}dx\]. . . . . . . . . . . . . . .(2)
\[I\]\[=\sin \left( \log x \right)\times x-\cos (\log x)\times x-\int{\sin \left( \log x \right)dx}\]
By equation(1) we substitute the value of $\int{\sin \left( \log x \right)dx}$ as $I$
\[I=\sin \left( \log x \right)\times x-\cos (\log x)\times x-I\]. . . . . . . . . . . . . . . . . . . . . . . . (3)
\[2I=x\left[ \sin \left( \log x \right)-\cos (\log x) \right]\]
\[I=\dfrac{x}{2}\left[ \sin \left( \log x \right)-\cos (\log x) \right]\]. . . . . . . . . . . . . . . . .. . (4)
The answer is option A.
Note: When they ask us to find the integration of a single function we can use normal integration formulas and if they ask us to find integration of product of two functions then we have to apply the integration by parts. If the two functions are different then we have to use the ILATE rule.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE