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CBSE Class 12-science Answered

-pi to pi (2x+2xsinx)/1+cos^2x
Asked by jritu1 | 05 Jan, 2016, 12:58: AM
answered-by-expert Expert Answer
I equals integral subscript negative pi end subscript superscript pi fraction numerator 2 x plus 2 x sin x over denominator 1 plus cos squared x end fraction d x R e p l a c e space x space b y space minus pi plus pi minus x space i n space t h e space i n t e g r a l space open curly brackets left parenthesis a plus b minus x right parenthesis space p r o p e r t y close curly brackets equals integral subscript negative pi end subscript superscript pi fraction numerator 2 open parentheses negative x close parentheses plus 2 open parentheses negative x close parentheses sin open parentheses negative x close parentheses over denominator 1 plus cos squared open parentheses negative x close parentheses end fraction rightwards double arrow I equals integral subscript negative pi end subscript superscript pi fraction numerator negative 2 x plus 2 x s i n x over denominator 1 plus c o s squared x end fraction d x therefore 2 I equals 4 integral subscript negative pi end subscript superscript pi fraction numerator x sin x over denominator 1 plus c o s squared x end fraction d x rightwards double arrow I equals 2 integral subscript negative pi end subscript superscript pi fraction numerator x s i n x over denominator 1 plus c o s squared x end fraction d x fraction numerator x s i n x over denominator 1 plus c o s squared x end fraction space i s space a n space e v e n space f u n c t i o n space sin c e space f left parenthesis x right parenthesis equals f left parenthesis negative x right parenthesis therefore integral subscript negative pi end subscript superscript pi fraction numerator x s i n x over denominator 1 plus c o s squared x end fraction d x equals 2 integral subscript 0 superscript pi fraction numerator x s i n x over denominator 1 plus c o s squared x end fraction d x therefore I equals 2 integral subscript negative pi end subscript superscript pi fraction numerator x s i n x over denominator 1 plus c o s squared x end fraction d x equals 4 integral subscript 0 superscript pi fraction numerator x s i n x over denominator 1 plus c o s squared x end fraction d x r e p l a c e space apostrophe x apostrophe space b y space apostrophe pi minus x apostrophe space i n space t h e space i n t e g r a l space open curly brackets a minus x space p r o p e r t y close curly brackets I equals 4 integral subscript 0 superscript pi fraction numerator open parentheses pi minus x close parentheses s i n open parentheses pi minus x close parentheses over denominator 1 plus c o s squared open parentheses pi minus x close parentheses end fraction d x equals 4 integral subscript 0 superscript pi fraction numerator open parentheses pi minus x close parentheses s i n x over denominator 1 plus c o s squared x end fraction d x 2 I equals 4 open square brackets integral subscript 0 superscript pi fraction numerator x s i n x over denominator 1 plus c o s squared x end fraction d x plus integral subscript 0 superscript pi fraction numerator open parentheses pi minus x close parentheses s i n x over denominator 1 plus c o s squared x end fraction d x close square brackets rightwards double arrow I equals 2 integral subscript 0 superscript pi fraction numerator pi s i n x over denominator 1 plus c o s squared x end fraction d x rightwards double arrow I equals 2 pi integral subscript 0 superscript pi fraction numerator s i n x over denominator 1 plus c o s squared x end fraction d x S u b s t i t u t e space cos x equals t minus sin x d x equals d t rightwards double arrow I equals 2 pi integral subscript 1 superscript negative 1 end superscript fraction numerator negative d t over denominator 1 plus t squared end fraction rightwards double arrow I equals negative 2 pi open square brackets tan to the power of negative 1 end exponent t close square brackets subscript 1 superscript negative 1 end superscript rightwards double arrow I equals negative 2 pi open parentheses tan to the power of negative 1 end exponent open parentheses negative 1 close parentheses minus tan to the power of negative 1 end exponent 1 close parentheses rightwards double arrow I equals negative 2 pi open parentheses negative t a n to the power of negative 1 end exponent 1 minus t a n to the power of negative 1 end exponent 1 close parentheses rightwards double arrow I equals 4 pi tan to the power of negative 1 end exponent 1
Answered by satyajit samal | 05 Jan, 2016, 12:37: PM

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