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MTH112: Calculus II-2025SPRG-LECT-02

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If  bold italic I subscript bold n bold equals bold integral( 7x + 7 )n cosx  dx  , Then  bold italic I subscript bold n bold plus49n(n-1) bold italic I subscript bold n bold minus bold 2 end subscript=.........

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Find a reduction formula for  I subscript n equals integral sin h to the power of n left parenthesis a x right parenthesis space d x

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Evaluate the following integral 

bold integral bold space fraction numerator bold t bold a bold n bold x over denominator square root of bold c bold o bold s bold x end root end fraction bold italic d bold italic x

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integral fraction numerator ln space x over denominator x end fraction d x

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integral fraction numerator x over denominator square root of 1 plus x squared end root end fraction d x equals square root of 1 plus x squared end root plus C

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integral fraction numerator x over denominator 1 plus x squared end fraction d x equals 0.5 space ln left parenthesis 1 plus x squared right parenthesis plus C

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integral x space e to the power of x squared end exponent space d x equals 0.5 space e to the power of x squared end exponent plus C space space

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Evaluate the following integral

integral fraction numerator e to the power of x over denominator 4 plus e to the power of 2 x end exponent end fraction d x

 

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Evaluate the following integral

integral fraction numerator e to the power of x over denominator square root of e to the power of 2 x end exponent minus 2 e to the power of x end root end fraction d x

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integral fraction numerator 1 over denominator square root of x open parentheses square root of x minus 1 close parentheses end fraction d xequals integral fraction numerator 2 over denominator u minus 1 end fraction space d u

What substitution did we make?

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