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Question Number 206536 by cortano21 last updated on 18/Apr/24

$$\text{Extra left or missing right}$$

Commented by Frix last updated on 18/Apr/24

$$\mathrm{Simply}\mathrm{use}t=\sqrt{1-2^x}$$

Answered by lepuissantcedricjunior last updated on 20/Apr/24

$$\mathit{I}=\int \frac{2^{\mathit{x}}\mathit{s}\mathit{i}\mathit{n}\left(\sqrt{4-2^{\mathit{x}+2}}\right)}{\sqrt{1-2^{\mathit{x}}}\mathit{c}\mathit{o}\mathit{s}\left(\sqrt{1-2^{\mathit{x}}}\right)}\mathit{d}\mathit{x}$$$$=\int \frac{2^{\mathit{x}}\mathit{s}\mathit{i}\mathit{n}\left(2\sqrt{1-2^{\mathit{x}}}\right)}{\sqrt{1-2^{\mathit{x}}}\mathit{c}\mathit{o}\mathit{s}\left(\sqrt{1-2^{\mathit{x}}}\right)}\mathit{d}\mathit{x}$$$$=\int \frac{2^{\mathit{x}+1}\mathit{s}\mathit{i}\mathit{n}\left(\sqrt{1-2^{\mathit{x}}}\right)}{\sqrt{1-2^{\mathit{x}}}}\mathit{d}\mathit{x}$$$$\mathit{p}\mathit{o}\mathit{s}\mathit{o}\mathit{n}\mathit{s}1-2^{\mathit{x}}={\mathit{t}}^2=>2^{\mathit{x}}=1-{\mathit{t}}^2$$$$=>\mathit{l}\mathit{n}2\times 2^{\mathit{x}}\mathit{d}\mathit{x}=-2\mathit{t}\mathit{d}\mathit{t}$$$$\Rightarrow \mathit{d}\mathit{x}=-\frac{2\mathit{t}}{(1-{\mathit{t}}^2)\mathit{l}\mathit{n}2}\mathit{d}\mathit{t}$$$$\iff \mathit{I}=\int \frac{-4\mathit{t}(1-{\mathit{t}}^2)\mathit{s}\mathit{i}\mathit{n}\mathit{t}}{\mathit{t}\mathit{l}\mathit{n}2(1-{\mathit{t}}^2)}\mathit{d}\mathit{t}$$$$=\int -4\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{t}}{\mathit{l}\mathit{n}\left(2\right)}\mathit{d}\mathit{t}=4\frac{\mathit{c}\mathit{o}\mathit{s}\mathit{t}}{\mathit{l}\mathit{n}2}+\mathit{c}\mathit{c}\in \mathbb{R}$$$$\iff \mathit{I}=\frac{4\mathit{c}\mathit{o}\mathit{s}\left(\sqrt{1-2^{\mathit{x}}}\right)}{\mathit{l}\mathit{n}2}+\mathit{c}\mathit{c}\in \mathbb{R}$$$$$$

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