If u - e ax cos bx dx and v - e ax sinbxdx- then a 2- b 2 u 2- v 2-A- 2 e axB- a2-b2 e2 a xC- e 2 axD- a2-b2 e2 a x

The correct option is C e^2axu =∫ e^ax cos bx dx =e^axsin bx/b a/b∫ e^ax. sin bx dx =e^axsin bx/b a/bv ⇒ bu +av =e^axsin bx ...(i) Similarly bv au= e^axcos bx ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

If u =∫ e ax ...

Question

If $u = \int e^{a x} c o s b x d x$ and $v = \int e^{a x} s i n b x d x$ , then $(a^{2} + b^{2}) (u^{2} + v^{2}) =$

2 e^{a x}

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(a^{2} + b^{2}) e^{2 a x}

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e^{2 a x}

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(a^{2} - b^{2}) e^{2 a x}

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Solution

The correct option is C $e^{2 a x}$
$u = \int e^{a x} c o s b x d x = e^{a x} \frac{s i n b x}{b} - \frac{a}{b} \int e^{a x} . s i n b x d x = \frac{e^{a x} s i n b x}{b} - \frac{a}{b} v \Rightarrow b u + a v = e^{a x} s i n b x . . . (i)$
Similarly $b v - a u = - e^{a x} c o s b x . . . . (i i)$
Squaring (i) and (ii) and adding, we get
$(a^{2} + b^{2}) (u^{2} + v^{2}) = e^{2 a x}$ .

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If u=∫eaxcos bx dx and v=∫eaxsinbxdx, then (a2+b2)(u2+v2)=

The correct option is C e2axu=∫eaxcos bx dx=eaxsinbxb−ab∫eax.sin bx dx=eaxsinbxb−abv⇒bu+av=eaxsin bx...(i) Similarly bv−au=−eaxcos bx....(ii) Squaring (i) and (ii) and adding, we get (a2+b2)(u2+v2)=e2ax.

If $u = \int e^{a x} c o s b x d x$ and $v = \int e^{a x} s i n b x d x$ , then $(a^{2} + b^{2}) (u^{2} + v^{2}) =$