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Maths-

General

Easy

Question

If $integral s e c to the power of fraction numerator 4 over denominator 9 end fraction end exponent invisible function application x times c o s invisible function application e to the power of fraction numerator 14 over denominator 9 end fraction end exponent x d x equals a t a n to the power of b end exponent invisible function application x plus c$ then a+b= ___

$\frac{106}{45}$
$\frac{45}{106}$
$- \frac{106}{45}$
$- \frac{9}{5}$

The correct answer is: $negative fraction numerator 106 over denominator 45 end fraction$

$integral fraction numerator s e c to the power of 2 end exponent invisible function application x d x over denominator left parenthesis t a n invisible function application x right parenthesis to the power of fraction numerator 14 over denominator 9 end fraction end exponent end fraction equals negative fraction numerator 9 over denominator 5 end fraction left parenthesis t a n invisible function application x right parenthesis to the power of negative fraction numerator 5 over denominator 9 end fraction end exponent plus c$

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The value of constant a>0 such that $stretchy integral subscript 0 end subscript superscript a end superscript open square brackets tan to the power of negative 1 end exponent invisible function application square root of x close square brackets d x equals stretchy integral subscript 0 end subscript superscript a end superscript open square brackets cot to the power of negative 1 end exponent invisible function application square root of x close square brackets d x$ where [.] denotes G.I.F is

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